Activities for First Order ODEs

Section 2.9 Activities for First Order ODEs

Subsection 2.9.1 Phase Lines

Activity 2.9.1.

Draw the phase line for this autonomous DE. (Assume this is a popluation, so that you only use positive values of \(p\text{.}\) You can use a computer to find approximate roots if necessary). Determine the stability of the solutions.

\begin{equation*} \frac{dp}{dt} = (p-3)(p-6)(p-10) \end{equation*}

Solution.

The roots of \((p-3)(p-6)(p-10)\) are \(3\text{,}\) \(6\) and \(10\) without any calculation, since the right side is already given in factored form. For \(p=2\text{,}\) the right side is negative, so the trajectory is on \((0,3)\) is downward. For \(p=5\text{,}\) the right side is positive, so the trajectory on \((3,6)\) is upward. For \(p=7\text{,}\) the right side is negative, so the trajectory is downward on \((6,10)\text{.}\) For \(p = 11\text{,}\) the right side is positive, so the trajectory is upward on \((10, \infty)\text{.}\) Figure 2.9.1 shows the phase line.

Figure 2.9.1. A Phase Line Diagram for \(\frac{dp}{dt} = (p-3)(p-6)(p-10)\)

Activity 2.9.2.

\begin{equation*} \frac{dp}{dt} = (p-2)^4(p-4)^7(p-6)^9 \end{equation*}

Solution.

The roots of \((p-2)^4(p-4)^7(p-6)^9\) are \(2\text{,}\) \(4\) and \(6\) without any calculation, since the right side is already in factored form. For \(p=1\text{,}\) the right side is positive, so the trajectory is upward on \((0,2)\text{.}\) For \(p=3\text{,}\) the right side is positive, so the trajectory is upward on \((2,4)\text{.}\) For \(p=5\text{,}\) the right side is negative, so the trajectory is downward on \((4,6)\text{.}\) for \(p=7\text{,}\) the right side is positive, so the trajectory is upward on \((6,\infty)\text{.}\) Figure 2.9.2 shows the phase line.

Figure 2.9.2. A Line Diagram for \(\frac{dp}{dt} = (p-2)^4(p-4)^7(p-6)^9\)

Activity 2.9.3.

\begin{equation*} \frac{dp}{dt} = \frac{1}{10} (p^4-15p^3+74p^2-138p+82) \end{equation*}

Solution.

The roots of \(\frac{1}{10} (p^4-15p^3+74p^2-138p+82)\) must be calculated by computer. Approximate values are \(p = 1.16\text{,}\) \(2.18\text{,}\) \(4.64\text{,}\) and \(7.03\text{.}\) There are four roots, so that are a non-repeated roots; therefore, the sign of the right side must change whever I cross a root. I start positive, so the trajectories are positive, negative, positive, negative, and positive. Figure 2.9.3 shows the phase line.

Figure 2.9.3. A Line Diagram for \(\frac{dp}{dt} = \frac{1}{10} (p^4-15p^3+74p^2-138p+82)\)

Activity 2.9.4.

\begin{equation*} \frac{dp}{dt} = (p-7)^2 \sinh (p-3) \end{equation*}

Solution.

The roots of \((p-7)^2 \sinh (p-3)\) are \(p=3\) and \(p=7\text{.}\) The right side evaluated at \(p=1\) is negative, so the trajectory on \((0,3)\) is downward. The right side evaluated at \(p=5\) is positive, so the trajectory on \((3,7)\) is upward. The right side evaluated at \(p=7\) is positive, so the trajectory on \((7, \infty)\) is upward. Figure 2.9.4 shows the phase line.

Figure 2.9.4. A Line Diagram for \(\frac{dp}{dt} = (p-7)^2 \sinh (p-3)\)

Activity 2.9.5.

\begin{equation*} \frac{dp}{dt} = \left( \ln \left( \frac{p}{5} \right) \right) \left( \cosh (p-2) - 1 \right) \end{equation*}

Solution.

The roots of \(\left( \ln \left( \frac{p}{5} \right) \right) \left( \cosh (x-2) - 1 \right)\) are \(p=2\) and \(p=5\text{.}\) The right side is negative for \(p=1\text{,}\) so the trajectory is downward on \((0,2)\text{.}\) The right side is negative for \(p=3\text{,}\) so the trajectory is downward on \((2,5)\text{.}\) The right side is positive for \(p=6\text{,}\) so the trajectory is upward on \((5,\infty)\text{.}\) Figure 2.9.5 shows the phase line.

Figure 2.9.5. A Line Diagram for \(\frac{dp}{dt} = \left( \ln \left( \frac{x}{5} \right) \right) \left( \cosh (x-2) - 1 \right)\)

Subsection 2.9.2 Direction Fields

Activity 2.9.6.

Draw a direction field for this DE. Describe the general form domain, and range of the solutions based on the direction field.

\begin{equation*} \frac{dy}{dx} = \frac{1}{x^2 + y^2 + 1} \end{equation*}

Solution.

The slope field for \(\frac{dy}{dx} = \frac{1}{x^2 + y^2 + 1}\) is in Figure 2.9.6. The solutions appear to be functions with slow growth between two horizontal asymptotes. They resemble arctangent functions. Their domains are \(\RR\) and their ranges seem to be fixed between the asymptotes.

Figure 2.9.6. A Slope Field for \(\frac{dy}{dx} = \frac{1}{x^2 + y^2 + 1}\)

Activity 2.9.7.

Draw a direction field for this DE. Describe the general form domain, and range of the solutions based on the direction field.

\begin{equation*} \frac{dy}{dx} = (x+4)(y-2) \end{equation*}

Solution.

The slope field for \(\frac{dy}{dx} = (x+4)(y-2)\) is in Figure 2.9.7. The function approach then diverge away from the line \(y=2\text{.}\) It’s not clear is they have vertical asymptotes or just fast growth. They might resemble quartic functions if they do not have asymptotes, where the vertex of the quartic is on the line \(x=-4\text{.}\) Their domains seem to be \(\RR\) and the ranges are likely bounded below or above, depending on a starting position. Note that \(y-2\) is also a singular solution that I can see in the slope field.

Figure 2.9.7. A Slope Field for \(\frac{dy}{dx} = (x+4)(y-2)\)

Activity 2.9.8.

Draw a direction field for this DE. Describe the general form domain, and range of the solutions based on the direction field.

\begin{equation*} \frac{dy}{dx} = \frac{x}{y^3} \end{equation*}

Solution.

The slope field for \(\frac{dy}{dx} = \frac{x}{y^3}\) is in Figure 2.9.8. First, the functions are undefined when \(y-0\text{,}\) so there is a family above the \(x\) axis and a family below the \(x\) axis. I can see from the slope field that these mirror each other. In the first and fourth quadrants, the functions diverge away from the \(x\) axis and look to approach horizontal asymptotes as \(x \rightarrow \infty\) (though this may simply be very slow growth). Their domains look like \((a, \infty)\text{,}\) where \(a\) is a starting point on the \(x\) axis, and their ranges seem to be \((0, \pm h)\text{,}\) where \(h\) is the \(y\) value of the asymptotes. For the second and third quadrants, the functions start from horizontal asymptotes and diverge toward the \(x\) axis. Their domains seem to be \((-\infty, -a)\text{,}\) where \(-a\) is a point on the \(x\) axis, and their ranges seem to be \((0, \pm h)\) where \(h\) is the height of the asymptote. If I connect the positive and negative pieces, I get horizontal quartic-like paths.

Figure 2.9.8. A Slope Field for \(\frac{dy}{dx} = \frac{x}{y^3}\)

Activity 2.9.9.

Draw a direction field for this DE. Describe the general form domain, and range of the solutions based on the direction field.

\begin{equation*} \frac{dy}{dx} = y \ln x \end{equation*}

Solution.

The slope field for \(\frac{dy}{dx} = y \ln x\) is in Figure 2.9.9. The solutions have domain \((0, \infty)\) to satisfy the \(\ln x\) inclusion in the equation. They are mirrored above and belod the \(x\) axis. They look to start descending steelply from the \(y\) axis, reach a max or min at \(y = \pm h\text{,}\) then diverge away from the axis again. They have vertical asymptotes at \(x=0\text{.}\) Their ranges seem to be \((\pm h, \pm \infty)\text{.}\) Note I can also see the \(y=0\) singular solution in the slope field.

Figure 2.9.9. A Slope Field for \(\frac{dy}{dx} = y \ln x\)

Activity 2.9.10.

Draw a direction field for this DE. Describe the general form domain, and range of the solutions based on the direction field.

\begin{equation*} \frac{dy}{dx} = x^2 + y^2 \end{equation*}

Solution.

The slope field for \(\frac{dy}{dx} = x^2 + y^2\) is in Figure 2.9.10. The solutions look to be steep function starting at one vertical asymptote and ending at another (unless, again, the asymptotes are just rapid growth). The function that passes through the origin looks either like a tangent function or a high power odd polynomial \(x^{2n+1}\text{.}\) The later is perhaps more likely, since there is a zero tangent at the origin. However, other function in the family lack this horizontal tangent. In any case, the range is clearly unbounded and the domain is either unbounded or fixed between the two vertical asymptotes.

Figure 2.9.10. A Slope Field for \(\frac{dy}{dx} = x^2 + y^2\)

Subsection 2.9.3 Separable Equations

Activity 2.9.11.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = x^2 y \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} \frac{1}{y} \frac{dy}{dx} = x^2 \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int \frac{1}{y} dy = \int x^2 dx \end{equation*}

These are both immediate integral: a logarithm integral and a power rule integral.

\begin{equation*} \ln |y| = \frac{x^3}{3} + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y = ce^{\frac{x^3}{3}} \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE.

Activity 2.9.12.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = \frac{\cos x}{e^y} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} e^y \frac{dy}{dx} = \cos x \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int e^y dy = \int \cos x dx \end{equation*}

These are immediate integrals from the tables of antiderivatives.

\begin{equation*} e^y = \sin x + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y = \ln (\sin x + c) \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE. Note that there are sone strict domain restriction for this function.

Activity 2.9.13.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = \frac{x^3}{y^3} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} y^3 \frac{dy}{dx} = x^3 \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int y^3 dy = \int x^3 dx \end{equation*}

These are both power rule integrals.

\begin{equation*} \frac{y^4}{4} = \frac{x^4}{4} + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y^4 = x^4 + c \implies y = \sqrt[4]{x^4 + c} \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE.

Activity 2.9.14.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = x^2 (y^2 +1) \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} \frac{1}{y^2 + 1} \frac{dy}{dx} = x^2 \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int \frac{1}{y^2 + 1} dy = \int x^2 dx \end{equation*}

The left is an arctangent integral, and the right is a power rule integral.

\begin{equation*} \arctan y = \frac{x^3}{3} + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y = \tan \left( \frac{x^3}{3} + c \right) \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE.

Activity 2.9.15.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = e^x \sqrt{4-y^2} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} \frac{1}{\sqrt{4-y^2}} \frac{dy}{dx} = e^x \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int \frac{1}{\sqrt{4-y^2}} dy = \int e^x dx \end{equation*}

The left is an arcsin integral, and the right is a simple exponential integral.

\begin{equation*} \arcsin \left( \frac{y}{2} \right) = e^x + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} \arcsin \left( \frac{y}{2} \right) = e^x + c \implies \frac{y}{2} = \sin (e^x + c) \implies y = 2 \sin (e^x + c) \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE.

Activity 2.9.16.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = \frac{y^2 -2y - 15}{x} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} \frac{1}{y^2 - 2y - 15} \frac{dy}{dx} = x \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int \frac{1}{y^2 - 2y = 15} = \int x dx \end{equation*}

The left is a partial fractions integral; I asked a computer for the solution. The right is a power rule integral.

\begin{equation*} \frac{1}{8} \ln |y-5| - \frac{1}{8} \ln |y-3| = x^2 + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\) However, I don’t see any way to pull the \(y\) out of both the logarithm terms. Therefore, I’ll leave the solution implicit.

Activity 2.9.17.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = \frac{x^4 + 9x^3 - 7x^2 - 3x + 1}{y} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} y \frac{dy}{dx} = x^4 + 9x^3 - 7x^2 + 3x + 1 \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int y dy = \int x^4 + 9x^3 - 7x^2 - 3x + 1 dx \end{equation*}

These are power polynomial integrals, solve by linearity and the power rule.

\begin{equation*} \frac{y^2}{2} = \frac{x^5}{5} + \frac{9x^4}{4} - \frac{7x^3}{3} - \frac{3x^2}{2} + x + c \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y = \sqrt{\frac{2x^5}{5} + \frac{9x^4}{2} - \frac{14x^3}{3} - 3x^2 + 2x + c} \end{equation*}

I was successful solving for \(y\) and producing a function which solves the DE.

Activity 2.9.18.

Solve this separable differential equation.

\begin{equation*} \frac{dy}{dx} = \frac{e^{-x^2}}{\cos y} \end{equation*}

Solution.

First, I bring the \(y\) expression to the left side of the equation.

\begin{equation*} \cos y \frac{dy}{dx} = e^{-x^2} \end{equation*}

Then I integrate. Using the standard substitution for separable equations, the left integral can be treated as an integral in \(y\text{.}\)

\begin{equation*} \int \cos y dy = \int e^{-x^2} dx \end{equation*}

This is a simple trig integral on the left. The right is an integral without an elementary solution. Therefore, I’ll just leave it as it is.

\begin{equation*} \sin y = \int e^{-x^2} ex \end{equation*}

After integrating, I try to solve for \(y\text{.}\)

\begin{equation*} y = \arcsin \left( \int e^{-x^2} dx \right) \end{equation*}

This is a solution, a function. It is defined by an integral, which makes it more difficult to work with and understand, but it is a solution.

Subsection 2.9.4 First Order Linear Equations

Activity 2.9.19.

Solve this linear equation.

\begin{equation*} \frac{dy}{dt} + y = 4e^{-t} \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = 1 \amp \amp Q(t) = 4e^{-t} \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int 1 dt} = e^t \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} e^t \frac{dy}{dt} + e^t y = 4e^{-t} e^t = 4 \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( e^t y \right) = 4 \end{equation*}

Then I integrate.

\begin{equation*} e^t y = 4t + c \end{equation*}

\begin{align*} y \amp = 4te^{-t} + ce^{-t} \\ y_p \amp = 4te^{-t} \\ y_h \amp = e^{-t} \end{align*}

Activity 2.9.20.

Solve this linear equation.

\begin{equation*} \frac{dy}{dt} + \frac{4y}{t} = t^2 + 3t + 4 \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = \frac{4}{t} \amp \amp Q(t) = t^2 + 3t + 4 \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int \frac{4}{t} dt} = e^{4 \ln t} = t^4 \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} t^4 \frac{dy}{dt} + t^4 \frac{4y}{t} = t^4(t^2 + 3t + 4) \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} (t^4 y) = t^6 + 3t^5 + 4t^4 \end{equation*}

Then I integrate.

\begin{equation*} t^4 y = \frac{t^7}{7} + \frac{t^6}{2} + \frac{4t^5}{5} + c \end{equation*}

\begin{align*} y \amp = \frac{t^3}{7} + \frac{t^2}{2} + \frac{4t}{5} + \frac{c}{t^4} \\ y_p \amp = \frac{t^3}{7} + \frac{t^2}{2} + \frac{4t}{5} \\ y_h \amp = \frac{1}{t^4} \end{align*}

Activity 2.9.21.

Solve this linear equation.

\begin{equation*} \frac{dy}{dt} + 3t^2y = t^2 \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = 3t^2 \amp \amp Q(t) = t^2 \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int 3t^3 dt} = e^{t^3} \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} e^{t^3} \frac{dy}{dt} + e^{t^3} 3t^2 y = e^{t^3} t^2 \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( e^{t^3} y \right) = e^{t^3}{t^2} \end{equation*}

Then I integrate. The right integral is a substitution integral with \(u = t^3\text{;}\) I haven’t shown the steps.

\begin{equation*} e^{t^3} y = \frac{e^{t^3}}{3} + c \end{equation*}

\begin{align*} y \amp = \frac{1}{3} + ce^{-t^3} \\ y_p \amp = \frac{1}{3} \\ y_h \amp = e^{-t^3} \end{align*}

Activity 2.9.22.

Solve this linear equation.

\begin{equation*} \frac{dy}{dt} + (\cos t) y = 4 \cos t \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = \cos t \amp \amp Q(t) = 4 \cos t \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int \cos t dt } e^{\sin t} \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} e^{\sin t} \frac{dy}{dt} + e^{\sin t} \cos t y = e^{\sin t} 4 \cos t \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( e^{\sin t} y \right) = e^{\sin t} 4 \cos t \end{equation*}

Then I integrate. The right side is a substitution with \(u = \sin y\text{,}\) but I haven’t shown the steps.

\begin{equation*} e^{\sin t} y = 4e^{\sin t} + c \end{equation*}

\begin{align*} y \amp = 4 + ce^{-\sin t} \\ y_p \amp = 4\\ y_h \amp = e^{-\sin t} \end{align*}

Activity 2.9.23.

Solve this linear equation. (Hint: when you get a hyperbolic inverse tangent, look up how to write it as a logarithm.)

\begin{equation*} \frac{dy}{dt} + \frac{2y}{1-t^2} = (1-t)^2 \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = \frac{2}{1-t^2} \amp \amp Q(t) = (1-t)^2 \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int \frac{2}{1-t^2} dt} = e^{2 \arc+anh t} \end{equation*}

I’m going to make use of a nice alternative definition of the hyperbolic arctangent.

\begin{equation*} \arctanh x = \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right) \implies \mu (t) = e^{2 \ln \left( \frac{1+t}{1-t} \right)} = \frac{1 + t}{1 - t} \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} \frac{1+t}{1-t} \frac{dy}{dt} + \frac{1+t}{1-t} \frac{y}{1-t^2} = \frac{1+t}{1-t} (1-t)^2 \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( \frac{1+t}{1-t} y \right) = (1+t)(1-t) = 1-t^2 \end{equation*}

Then I integrate.

\begin{equation*} \frac{1+t}{1-t} y = t - t^3 + c \end{equation*}

\begin{align*} y \amp = (t-t^3) \frac{1-t}{1+t} + c \frac{1-t}{1+t} \\ y_p \amp = \frac{(t-t^3)(1-t)}{1+t} \\ y_h \amp = \frac{1-t}{1+t} \end{align*}

Activity 2.9.24.

Solve this linear equation.

\begin{equation*} \frac{dy}{dt} + (\tan t) y = e^t(t^2 -1) \end{equation*}

Solution.

I follow the standard procedure. First I’ll label the coefficient functions.

\begin{align*} \amp P(t) = \tan t \amp \amp Q(t) = e^t (t^2 - 1) \end{align*}

Then I’ll calculate the integrating factor. (Note that I don’t need a constant of integration for the integrating factor).

\begin{equation*} \mu(t) = e^{\int \tan t dt} = e^{-\ln (\cos t)} = (\cos t)^{-1} = \sec t \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} \sec t \frac{dy}{dt} + (\sec t)(\tan t) y = \sec t e^t (t^2 - 1) \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( y \sec t \right) = \sec t e^t (t^2 - 1) \end{equation*}

Then I integrate.

\begin{equation*} y \sec t = \int \sec t e^t (t^2 - 1) dt + c \end{equation*}

Then I solve for \(y\text{.}\) If I want to, I can isolate the particular solution \(y_p\) and the homogeneous solution \(y_h\text{;}\) the homogeneous solution has the paramater and the particular solution does not. Here, I’m going to leave the integral as is on the right side and expression the solution in terms of that integral.

\begin{align*} y \amp = \cos t \int \sec t e^t (t^2 - 1) dt + c \cos t \\ y_p \amp = \cos t \int \sec t e^t (t^2 - 1) dt \\ y_h \amp = \cos t \end{align*}

Subsection 2.9.5 Substitution - Homogeneous Equations

Activity 2.9.25.

Solve this homogeneous equation using substitution.

\begin{equation*} \frac{dy}{dt} = \frac{y^3}{t^3} - 1 \end{equation*}

Solution.

I use the appropriate substitution: \(v = \frac{y}{t}\text{.}\) The expression on the left becomes \(f(v) = v^3 - 1\text{.}\) Then I can write the new differential equation in \(v\text{.}\)

\begin{equation*} \frac{dv}{dt} = \frac{f(v) - v}{t} = \frac{v^3 - 1 - v}{t} \end{equation*}

As expected, this is a separable equation. I proceed with the steps to solve a separable equation.

\begin{equation*} \frac{1}{v^3 - v - 1} \frac{dv}{dt} = \frac{1}{t} \end{equation*}

Then I integrate with the standard setup that turns the left side into a integral in \(v\text{.}\)

\begin{equation*} \int \frac{1}{v^3 - v - 1} dv = \int \frac{1}{t} dt \end{equation*}

The right side is an easy power rule integral. On the left side, I need partial fractions. I’ve asked a computer for the partial fraction decomposition and the outcome of the resulting integrals.

\begin{equation*} \frac{1}{11} \ln |v-2| - \frac{1}{22} \ln |v^2 + 2v + 3| - \frac{3}{11\sqrt{2}} \arctan \left( \frac{v+1}{\sqrt{2}} \right) = \ln |t| + c \end{equation*}

Solving on the left for \(v\) is clearly impossible, so I’ll have to leave this as an implicit solution. I can still replace the original substitution, leaving an even more complicated implicit expression for the function \(y(t)\text{.}\)

\begin{equation*} \frac{1}{11} \ln \left| \frac{y}{t} - 2 \right| - \frac{1}{22} \ln \left | \frac{y^2}{t^2} + \frac{2y}{t} + 3 \right| - \frac{3}{11\sqrt{2}} \arctan \left( \frac{\frac{y}{t}+1}{\sqrt{2}} \right) = \ln |t| + c \end{equation*}

Activity 2.9.26.

Solve this homogeneous equation using substitution.

\begin{equation*} \frac{dy}{dt} = \frac{t}{t - y} \end{equation*}

Solution.

This is a homogeneous equation, but it is a bit difficult to immediately see it. However, if I divide numerator and denominator by \(t\) on the right, the form will appear.

\begin{equation*} \frac{dy}{dt} = \frac{1}{1- \frac{y}{t}} \end{equation*}

I use the appropriate substitution: \(v = \frac{y}{t}\text{.}\) The expression on the left becomes \(f(v) = \frac{1}{1-v}\text{.}\) Then I can write the new differential equation in \(v\text{.}\)

\begin{align*} \frac{dv}{dt} \amp = \amp \frac{f(v) - v}{t} = \frac{\frac{1}{1-v} - v}{t} = \frac{\frac{1 - v(1-v)}{1-v}}{t} \\ \amp = \amp \frac{1 - v + v^2}{t(1-v)} \end{align*}

As expected, this is a separable equation. I solve it as a separable equation, setting up the two integrals. in \(t\)

\begin{equation*} \int \frac{1 - v}{1 - v + v^2} dv = \int \frac{1}{t} dt \end{equation*}

Then I did the integrals (asking a computer for the left integral.

\begin{equation*} \frac{1}{\sqrt{3}} \arctan \left( \frac{2v-1}{\sqrt{3}} \right) - \frac{1}{2} \ln |v^2 - v + 1| = \ln |t| + c \end{equation*}

In principle, I would like to solve this for v. However, this turns out to be impossible: there is no way to extra v from both the arctangent and the logarithm (as well as the quadratic inside the logarithm). The best I can do is replace \(v\) with \(\frac{y}{t}\) and leave the solution in an implicit form.

\begin{equation*} \frac{1}{\sqrt{3}} \arctan \left( \frac{\frac{2y}{t}-1}{\sqrt{3}} \right) - \frac{1}{2} \ln \left| \left( \frac{y}{t} \right)^2 - \frac{y}{v} + 1 \right| = \ln |t| + c \end{equation*}

Subsection 2.9.6 Substitution - Bernoulli Equations

Activity 2.9.27.

Solve this Bernoulli equation by substitution.

\begin{equation*} \frac{dy}{dt} + \frac{y}{t} = \frac{y^3}{t^2} \end{equation*}

Solution.

In this Bernoulli equation, \(n=3\text{,}\) so the substitution is \(v = y^{-2}\text{.}\) The substitution produces a linear equation.

\begin{equation*} \frac{dv}{dt} - \frac{2v}{t} = - \frac{2}{t^2} \end{equation*}

I proceed with the usual steps for linear equations. I identify the two coefficient functions.

\begin{align*} \amp P(t) = \frac{-2}{t} \amp \amp Q(t) = \frac{-2}{t^2} \end{align*}

Then I’ll calculate the integrating factor.

\begin{equation*} \mu(t) = e^{\int \frac{-2}{t} dt} = e^{-2\ln |t|} = \frac{1}{t^2} \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} \frac{1}{t^2} \frac{dv}{dt} - \frac{2v}{t^3} = - \frac{2}{t^4} \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( \frac{1}{t^2} v \right) = - \frac{2}{t^4} \end{equation*}

Then I integrate.

\begin{equation*} \frac{1}{t^2} v = \frac{2}{3t^3} + c \end{equation*}

Then I solve for \(v\text{.}\)

\begin{equation*} v = \frac{2}{3t} + ct^2 \end{equation*}

Then I can reverse the substitution.

\begin{equation*} \frac{1}{y^2} = \frac{2}{3t} + ct^2 = \frac{2 + ct^3}{3t} \end{equation*}

Finally, I can solve for \(y\)

\begin{equation*} y = \sqrt{\frac{3t}{2 + ct^3}} \end{equation*}

Activity 2.9.28.

Solve this Bernoulli equation by substitution.

\begin{equation*} \frac{dy}{dt} + \sin t y = t^2 y^4 \end{equation*}

Solution.

In this Bernoulli equation, \(n=4\text{,}\) so the substitution is \(v = y^{-3}\text{.}\) The substitution produces a linear equation.

\begin{equation*} \frac{dv}{dt} -3 \sin t v = -3t^2 \end{equation*}

I proceed with the usual steps for linear equations. I identify the two coefficient functions.

\begin{align*} \amp P(t) = -3 \sin t \amp \amp Q(t) = -3t^2 \end{align*}

Then I’ll calculate the integrating factor.

\begin{equation*} \mu(t) = e^{\int -3 \sin t dt} = e^{3 \cos t} = \end{equation*}

Then I multiply the original equation by the integrating factor.

\begin{equation*} e^{3 \cos t} \frac{dv}{dt} - 3e^{3 \cos t} \sin t v = -3e^{3 \cos t} t^2 \end{equation*}

The left side becomes a product rule derivative.

\begin{equation*} \frac{d}{dt} \left( e^{3 \cos t} v \right) = -3e^{3 \cos t} t^2 \end{equation*}

Then I integrate.

\begin{equation*} v e^{3 \cos t} = \int -3e^{3 \cos t} t^2 dt \end{equation*}

I have to leave the integral on the right, since I don’t know any way to solve it with the techniques of elementary functions. Likely, this solution to the integral is some previously unknown function. However, I can solve for \(v\text{.}\)

\begin{equation*} v = e^{-3 \cos t} \int -3e^{3 \cos t} t^2 dt \end{equation*}

Then I can reverse the substitution.

\begin{equation*} \frac{1}{y^3} = e^{-3 \cos t} \int -3e^{3 \cos t} t^2 dt \end{equation*}

Finally, I can solve for \(y\)

\begin{equation*} y = \sqrt[3]{\frac{1}{e^{-3 \cos t} \int -3e^{3 \cos t} t^2 dt}} \end{equation*}

This is an unwieldy solution; it would be quite difficult to investigate the properties of this strangely defined function. But it is a solution.

Prev Top Next