Week 9 Activity

Section 9.4 Week 9 Activity

Subsection 9.4.1 Substitution Rule

Activity 9.4.1.

Calculate this integral.

\begin{equation*} \int 3x^2 (x^3 - 4)^5 dx \end{equation*}

Solution.

The second part of this integrand looks like a composition: \((x^3 - 4)^5\) is an inside function, with an outside function \(u^5\text{.}\) Moreover, the remaining piece of the integral looks like it might be the derivative of the inside, which is idea for the substitution rule. Here is the substitution.

\begin{align*} u \amp = x^3 - 4 \\ du \amp = 3x^2 dx \end{align*}

I can rearrange the original integral to match these pieces, then do the replacement.

\begin{align*} \int 3x^2 (x^3 - 4)^5 dx \amp = \int (x^3-4)^4 (3x^2 dx) \\ \amp = \int u^5 du = \frac{u^6}{6} + C \end{align*}

At the end, I undo the substitution to return to the original variable.

\begin{equation*} \frac{u^6}{6} + C = \frac{(x^3-4)^6}{6} + C \end{equation*}

Activity 9.4.2.

Calculate this integral.

\begin{equation*} \int \cos (17 x) dx \end{equation*}

Solution.

The integrand is a composition with inside \((17x)\) and outside \(\cos (u)\text{.}\) We’ll try a substitution with these pieces.

\begin{align*} u \amp = 17 x \\ du \amp = 17 dx \\ \frac{1}{17} du \amp = dx \end{align*}

I had to move the 17 to the left, since the original integral doesn’t have a term which looks like \(17 dx\text{.}\) In this form, however, I can replace all the piece of the original integral with the new variable.

\begin{equation*} \int \cos (17 x) dx = \int \cos u \frac{1}{17} du \end{equation*}

Then I can use linearity and known antideriatives to do the integral.

\begin{equation*} = \frac{1}{17} \sin (u) + C \end{equation*}

Finally, I reverse the substitution.

\begin{equation*} = \frac{1}{17} \sin (17 x) + C \end{equation*}

Activity 9.4.3.

Calculate this integral.

\begin{equation*} \int \frac{x-4}{x-5} dx \end{equation*}

Solution.

This doesn’t look like a composition. However, substitution is sometime still useful even when I don’t obviously have a composition. Here, I’m going to take the denominator as the substitution, to try to produce an easier integral.

\begin{align*} u \amp = x - 5\\ du \amp = dx \end{align*}

I also need to know how to replace the numerator.

\begin{equation*} u = x - 5 \implies u + 1 = x - 4 \end{equation*}

Now I can replace all the piece of the original integral.

\begin{equation*} \int \frac{x-4}{x-5} dx = \int \frac{u+1}{u} du \end{equation*}

Now that I have a simpler denominator, I can split the numerator up to give two integrals, using linearity.

\begin{equation*} = \int \frac{u}{u} + \frac{1}{u} du = \int 1 du + \int \frac{1}{u} du \end{equation*}

Then I use known antiderivatives.

\begin{equation*} = u + \ln |u| + C \end{equation*}

Finally, I reverse the substitution.

\begin{equation*} = (x-5) + \ln |x-5| + C \end{equation*}

As a final note, I could group the \((-5)\) with the constant, like this:

\begin{equation*} = x + \ln |x-5| + (C - 5) \end{equation*}

If I did that, since the constant can be set to any real number, subtracting 5 doesn’t make a difference. Therefore, an alternative form of the answer gets ride of this \((-5)\text{.}\)

\begin{equation*} = x + \ln |x-5| + C \end{equation*}

This certinaly isn’t necessary, but I mention it here since these kinds of steps with constants of integration are commonly done.

Activity 9.4.4.

Calculate this integral.

\begin{equation*} \int_0^1 2x(x^2 + 4)^3 dx \end{equation*}

Solution.

This looks like a composition with inside function \((x^2+4)\) and outside function \(u^3\text{.}\) The rest of the integrand looks like it might be close to the derivative of the inside function, which is promising. We proceed with the substitution rule. Now I will also change the bounds.

\begin{align*} u \amp = x^2 + 4\\ du \amp = 2x dx\\ x \amp = 0 \implies u = (0)^2 + 4 = 4 \\ x \amp = 1 \implies u = (1)^2 + 4 = 5 \end{align*}

I can rearrange the integral so that I can substitute all the pieces clearly.

\begin{equation*} \int_0^1 2x(x^2 + 4)^3 dx = \int_0^1 (x^2+4)^3 (2x dx) \end{equation*}

Then I replace all the pieces, including changing the bounds.

\begin{equation*} = \int_4^5 u^3 du \end{equation*}

Then I can proceed to finish the integral. Since this is a definite integral and I have changed the bounds as well, I do not need to reverse the substitution.

\begin{equation*} = \frac{u^4}{4} \Big|_4^5 = \frac{5^4}{4} - \frac{4^4}{4} = \frac{625 - 256}{4} = \frac{369}{4} \end{equation*}

Activity 9.4.5.

Calculate this integral.

\begin{equation*} \int_0^4 \frac{x}{x+3} dx \end{equation*}

Solution.

I’d like to simplify the denominator, so I’ll try a substitution with \(u = (x+3)\text{.}\) Since this is a definite integral, I’ll change the bounds as well. Finally, I’ll need to replace the numerator in the substitution, so I’ll figure out how to express the numerator in the new variable.

\begin{align*} u \amp = x+3\\ du \amp = dx \\ x \amp = 0 \implies u = 0 + 3 = 3 \\ x \amp = 4 \implies u = 4 + 3 = 7 \\ u \amp = x + 3 \implies x = u - 3 \end{align*}

Now I have all the pieces and I can change to the new variable, including changing the bounds.

\begin{equation*} \int_0^4 \frac{x}{x+3} dx = \int_3^7 \frac{u-3}{u} du \end{equation*}

Then I can split up the integral using linearity into two integrals that I can solve with known antiderivatives.

\begin{align*} \amp = \int_3^7 \frac{u}{u} du - 3 \int_3^7 \frac{1}{u} du \\ \amp = \int_3^7 1 du - 3 \int_3^7 \frac{1}{u} du \\ \amp = u \Big|_3^7 - 3 \ln |u| \Big|_3^7 = (7-3) - 3 (\ln 7 - \ln 3) = 4 - 3 (\ln 7 - \ln 3) \end{align*}

Activity 9.4.6.

Calculate this integral.

\begin{equation*} \int_0^4 \frac{x}{x-3} dx \end{equation*}

Solution.

This looks very similar to the previous question. One might thing that the process should be exactly the same. You could follow that process and get an answer, but that answer would not be valid. The problem, here, is that this is a definite integral, which calculates an area under the curve. Given these bounds, I am looking for the area between 0 and 4. However, this function is undefined at \(x=3\text{,}\) where it has a vertical asymptote. Therefore, the notion of area under this curve is meaningless. There is no connected curve under which I can define an area. This integral has no answer.

Subsection 9.4.2 Initial Value Problems

Activity 9.4.7.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = x^2 y^2 \amp \amp y(0) = 1 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = x^2\) and \(g(y) = y^2\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

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

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

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

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 1 = \frac{-3}{x + 3c} \implies c = -1 \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

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

Activity 9.4.8.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = e^x e^{-y} \amp \amp y(0) = 1 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = e^x\) and \(g(y) = e^{-y}\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

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

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

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

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 1 = \ln (e^0 + c) = \ln (1+c) \implies c = e-1 \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = \ln (e^x + e - 1) \end{equation*}

Activity 9.4.9.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = \frac{y}{x} \amp \amp y(3) = 1 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = \frac{1}{x}\) and \(g(y) = y\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

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

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

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

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 1 = 3(e^c) \implies (e^c) = \frac{1}{3} \implies c = \ln \frac{1}{3} = -\ln 3 \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = x e^{-\ln 3} = \frac{x}{e^{\ln 3}} = \frac{x}{3} \end{equation*}

Activity 9.4.10.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = \sqrt{\frac{x}{y}} \amp \amp y(0) = 0 \end{align*}

Solution.

The expression in the differential equation is not defined for this initial value. Therefore, this IVP cannot be solved. I could do the separable equation, but I won’t be able to input the initial conditions.

Activity 9.4.11.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = y^2 \sin x \amp \amp y(\pi) = 1 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = \sin x^2\) and \(g(y) = y^2\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

\begin{equation*} \frac{-1}{y} = -\cos x + c \end{equation*}

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

\begin{equation*} y = \frac{-1}{c - \cos x} = \frac{1}{\cos x - c} \end{equation*}

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 1 = \frac{1}{\cos \pi - c} = \frac{1}{-1 - c} \implies c= -2 \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = \frac{1}{\cos x + 2} \end{equation*}

Activity 9.4.12.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = \frac{(x-1)^2}{y} \amp \amp y(2) = 1 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = (x-1)2\) and \(g(y) = \frac{1}{y}\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

\begin{equation*} \frac{y^2}{2} = \frac{(x-1)^3}{3} + c \end{equation*}

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

\begin{equation*} \frac{y^2}{2} = \frac{(x-1)^3+3c}{3} \implies y = \sqrt{\frac{2(x-1)^3+6c}{3}} \end{equation*}

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 1 = \sqrt{\frac{2(2-1)^3+6c}{3}} \implies 3 = 2+6c \implies c = \frac{1}{6} \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = \sqrt{ \frac{2(x-1)^3+1}{3}} \end{equation*}

Activity 9.4.13.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = x^3 y^4 \amp \amp y(1) = -3 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = x^3\) and \(g(y) = y^4\text{.}\)

\begin{equation*} \int \frac{1}{y^4} dy = \int x^3 dx \end{equation*}

Then I integrate both sides, add the constant of integration on the right.

\begin{equation*} \frac{-1}{3y^3} = \frac{x^4}{4} + c = \frac{x^4 + 4c}{4} \end{equation*}

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

\begin{equation*} -3y^3 = \frac{4}{x^4 + 4c} \implies y^3 = \frac{-4}{3(x^4+4c)} \implies y = \sqrt[3]{\frac{-4}{3x^4 + 12c}} \end{equation*}

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} -3 = \sqrt[3]{\frac{-4}{3(1)^4+ 12c}} \implies -27 = \frac{-4}{3+12c} \end{equation*}

\begin{equation*} \frac{1}{27} = \frac{3+12c}{4} \implies 3+12c = \frac{4}{27} \implies 12c = \frac{-77}{27} \implies c = \frac{-77}{324} = \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = \sqrt[3]{\frac{-4}{3x^4 + 12\frac{-77}{324}}} = \sqrt[3]{\frac{-4}{3x^2 - \frac{77}{27}}} = \sqrt{3}{\frac{-108}{81x^2 - 77}} = \sqrt{3}{\frac{108}{77 - 81x^2}} \end{equation*}

Activity 9.4.14.

Solve this initial value problem.

\begin{align*} \frac{dy}{dx} \amp = y^2 e^x \amp \amp y(\ln 3) = 2 \end{align*}

Solution.

I use the form for separable equations, with \(f(x) = e^x\) and \(g(y) = y^2\text{.}\)

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

Then I integrate both sides, add the constant of integration on the right.

\begin{equation*} \frac{-1}{y} = e^x + c \end{equation*}

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

\begin{equation*} y = \frac{-1}{e^x + c} \end{equation*}

Now I put the initial condition values in for \(x\) and \(y\text{,}\) to try to solve for the constant of integration.

\begin{equation*} 2 = \frac{-1}{3+c} \implies c= \frac{-7}{2} \end{equation*}

Finally, having solve for the constant of integration, I replace \(c\) with the value solved for to get the final, unique solution.

\begin{equation*} y = \frac{-1}{e^x - \frac{7}{2}} = \frac{2}{7 - 2e^x} \end{equation*}

Subsection 9.4.3 Exponential Growth

Activity 9.4.15.

Solve this initial value problem, where \(p(t)\) is population in terms of time.

\begin{align*} \frac{dp}{dt} \amp = (0.03) p(t) \\ p(0) \amp = 14000 \end{align*}

Solution.

I follow the process for solving separable equations, using \(g(p) = p\) and \(f(t) = 0.03\text{.}\)

\begin{equation*} \int \frac{1}{p} dp = \int 0.03 dt \end{equation*}

I integate both sides; both a direct antiderivatives.

\begin{equation*} \ln |p| = 0.03t + c \end{equation*}

Then I solve for \(p\text{.}\) I can drop the absolute value since population must be positive.

\begin{equation*} p(t) = e^{(0.03t) + c} = e^c e^{0.03t} \end{equation*}

Finally, I use the initial value p(0) = 14000.

\begin{equation*} p(0) = e^c e^0 = e^c = 14000 \end{equation*}

Here is the final solution.

\begin{equation*} p(t) = 14000 e^{0.03t} \end{equation*}

This is exactly what I expected: population that starts at 14000 with a growth rate coefficient of \(0.03\text{.}\)

Activity 9.4.16.

Solve this initial value problem, where \(p(t)\) is population in terms of time.

\begin{align*} \frac{dp}{dt} \amp = (-0.14) p(t) \\ p(0) \amp = 150000 \end{align*}

Solution.

I follow the process for solving separable equations, using \(g(p) = p\) and \(f(t) = 0.03\text{.}\)

\begin{equation*} \int \frac{1}{p} dp = \int -0.14 dt \end{equation*}

I integate both sides; both a direct antiderivatives.

\begin{equation*} \ln |p| = (-0.14t) + c \end{equation*}

Then I solve for \(p\text{.}\) I can drop the absolute value since population must be positive.

\begin{equation*} p(t) = e^c e^{-0.14t} \end{equation*}

Finally, I use the initial value p(0) = 14000.

\begin{equation*} p(0) = e^c e^0 = e^c = 1500000 \end{equation*}

Here is the final solution.

\begin{equation*} p(t) = 150000 e^{-0.14t} \end{equation*}

This is exactly what I expect: population that starts at 150000 with a growth rate coefficient of \(-0.14\text{.}\)

Subsection 9.4.4 Logistic Growth

Activity 9.4.17.

Check that the logistic growth function given solves the initial value problem. \(p(t)\) is population in terms of time. (First check the initial value. Then calculate the right side and left side of the differential equation and see if you can simpify both expression into the same thing.)

\begin{align*} p(t) \amp = \frac{400000 e^{0.05t}}{600 + 400 e^{0.05t}} \\ \frac{dp}{dt} \amp = (0.05) p(t) \left( 1 - \frac{p(t)}{1000} \right) \\ p(0) \amp = 400 \end{align*}

Solution.

I have to check two things. First, I can check that the initial value is correct. This is simply evaluating \(p(0)\text{.}\)

\begin{equation*} p(0) = \frac{400000 e^0}{600 + 400 e^0} = \frac{400000}{600+ 400} = \frac{400000}{1000} = 400 \end{equation*}

This is the correct initial value. Now I need to differentiate \(p(t)\) using the quotient rule and check that it is the same as the left side of the differential equation. This calculation is a bit complicated with many pieces, but I’ll simplify it step by step.

\begin{align*} \frac{dp}{dt} \amp = \frac{(600 + 400e^{0.05t}) \left( \frac{d}{dt} 400000 e^{0.05t} \right) - 400000 e^{0.05t} \left( \frac{d}{dt} 600 + 400 e^{0.05t} \right)}{(600 + 400 e^{0.05t})^2}\\ \amp = \frac{(600 + 400e^{0.05t})((0.05)(400000)e^{0.05t} - 400000 e^{0.05t}((0.05)400e^{0.05t})}{(600 + 400 e^{0.05t})^2}\\ \amp = \frac{(600 + 400e^{0.05t})20000 e^{0.05t} - 400000 e^{0.05t} 20 e^{0.05t}}{(600 + 400 e^{0.05t})^2}\\ \amp = \frac{12000000e^{0.05t} + 8000000e^{0.1t} - 8000000 e^{0.1t}}{(600 + 400 e^{0.05t})^2}\\ \amp = \frac{12000000e^{0.05t}}{(600 + 400 e^{0.05t})^2} \end{align*}

After all that simplification, I have this expression. This is the left side of the differential equation. I want to compare it to the right side of the differential equation. I’ll try to calculate that right side and see if I can simplify it down to the same thing. From the third to the fourth line below, I construct a common denominator for the term in brackets.

\begin{align*} \amp (0.05) p(t) \left( 1 - \frac{p(t)}{1000} \right) \\ \amp = (0.05) \frac{400000 e^{0.05t}}{600 + 400 e^{0.05t}} \left( 1 - \frac{ \frac{400000 e^{0.05t}}{600 + 400 e^{0.05t}}}{1000} \right)\\ \amp = \frac{20000 e^{0.05t}}{600 + 400 e^{0.05t}} \left( 1 - \frac{400 e^{0.05t}}{600 + 400 e^{0.05t}} \right)\\ \amp = \frac{20000 e^{0.05t}}{600 + 400 e^{0.05t}} \left( \frac{600 + 400 e^{0.05t} - 400 e^{0.05t}}{600 + 400 e^{0.05t}} \right)\\ \amp = \frac{20000 e^{0.05t}}{600 + 400 e^{0.05t}} \left( \frac{600}{600 + 400 e^{0.05t}} \right)\\ \amp = \frac{12000000 e^{0.05t}}{(600 + 400 e^{0.05t})^2} \end{align*}

This produces exactly the same expression. Both the left and ride sides of the equation simplify to the same expression. Therefore, since these simplification have made any actual changes to the function (just to how it is expressed), the right side and left side of the differential equation are equal, and I have confirmed that the given function solves the differential equation.

Activity 9.4.18.

Check that the logistic growth function given solves the initial value problem. \(p(t)\) is population in terms of time.

\begin{gather*} p(t) = \frac{240000 e^{0.02t}}{-500 + 800 e^{0.02t}} \\ \frac{dp}{dt} = (0.02) p(t) \left( 1 - \frac{p(t)}{300} \right) \\ p(0) = 800 \end{gather*}

Solution.

I have to check two things. First, I can check that the initial value is correct. This is simply evaluating \(p(0)\text{.}\)

\begin{equation*} p(0) = \frac{240000 e^0}{-500 + 800 e^0} = \frac{240000}{-500+ 800} = \frac{240000}{300} = 800 \end{equation*}

\begin{align*} \frac{dp}{dt} \amp = \frac{(-500 + 800e^{0.02t}) \left( \frac{d}{dt} 240000 e^{0.02t} \right) - 240000 e^{0.02t} \left(\frac{d}{dt} \left( -500 + 800 e^{0.02t} \right) \right)}{(-500 + 800 e^{0.02t})^2}\\ \amp = \frac{(-500 + 800e^{0.02t})(0.02)(240000)e^{0.02t} - 240000 e^{0.02t}((0.02)(800)e^{0.02t})}{(-500 + 800 e^{0.02t})^2}\\ \amp = \frac{(-500 + 800e^{0.02t})(4800)e^{0.02t} - 240000 e^{0.02t}((16)e^{0.02t})}{(-500 + 800 e^{0.02t})^2}\\ \amp = \frac{-2400000e^{0.02t} - 3840000e^{0.04t} + 3840000 e^{0.04t}}{(-500 + 800 e^{0.02t})^2}\\ \amp = \frac{-2400000e^{0.02t}}{(-500 + 800 e^{0.02t})^2} \end{align*}

After all that simplification, I have this expression. This is the left side of the differential equation. I want to compare it to the right side of the differential equation. Let’s try to calculate that right side and see if I can simplify it down to the same thing. From the third to the fourth line below, I construct a common denominator for the term in brackets.

\begin{align*} \amp (0.02) p(t) \left( 1 - \frac{p(t)}{300} \right) \\ \amp = (0.02) \frac{240000 e^{0.02t}}{-500 + 800 e^{0.02t}} \left( 1 - \frac{ \frac{240000 e^{0.02t}}{-500 + 800 e^{0.02t}}}{300} \right)\\ \amp = \frac{4800 e^{0.02t}}{-500 + 800 e^{0.02t}} \left( 1 - \frac{800 e^{0.02t}}{-500 + 800 e^{0.02t}} \right)\\ \amp = \frac{4800 e^{0.02t}}{-500 + 800 e^{0.02t}} \left( \frac{-500 + 800e^{0.02t} - 800 e^{0.02t}}{-500 + 800 e^{0.02t}} \right)\\ \amp = \frac{4800 e^{0.02t}}{-500 + 800 e^{0.02t}} \left( \frac{-500}{-500 + 800 e^{0.02t}} \right)\\ \amp = \frac{-24000000 e^{0.02t}}{(-500 + 800 e^{0.02t})^2} \end{align*}

Subsection 9.4.5 Conceptual Review Questions

How is the substitution rule an inverse to the chain rule?
What does the substitution rule mean about areas under curves?
How are exponential growth and logistic growth encoded in differential equations?
What does it mean to solve a differential equation?
What is an initial condition and why do I need them to solve differential conditions?

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