Question

Differential Equations.
Method of Variation question.

Answer 1

First you need some known solutions to be able to use variation of parameters. To get these, you solve for the homogeneous solutions:
\[y''+y=0\\
y''+3y'+2y=0\]

Answer 2

For the first one, you get a characteristic polynomial
\[r^2+1=0\]
which has roots \(\pm i\). This gives you a general homogeneos solution \(y=C_1\cos x+C_2\sin x\). For the VoP formulas, you would set \(y_1=\cos x\) and \(y_2=\sin x\).

The solutions from VoP have the form \(y=u_1y_1+u_2y_2\).

The formulas for the \(u\)'s are:
\[u_1=-\int\frac{y_2g(x)}{W(y_1y_2)~dx}\\
u_2=\int\frac{y_1g(x)}{W(y_1y_2)~dx}\]
where \(g(x)=\cos^2x\) in this case. \(W(a,b)\) is the Wronskian of \(a\) and \(b\).

Answer 3

\[W(y_1,y_2)=\begin{vmatrix}\cos x&\sin x\\-\sin x&\cos x\end{vmatrix}=\cos^2x+\sin^2x=1\]
Because the Wronskian is non-zero the solutions \(y_1\) and \(y_2\) are linearly independent, so you can proceed.
\[u_1=-\int\sin x\cos^2x~dx\\
u_2=\int \cos x\cos^2x~dx\]
Not too hard, right?

Answer 4

yes i have gotten here this where my problem is

Answer 5

So you just have trouble integrating?

Answer 6

yes

Answer 7

i have tried many identities not sure im doing right

Answer 8

The first one is a simple substitution, \(t=\cos x\) so that \(-dt=\sin x~dx\), then
\[\begin{align*}u_1&=-\int\sin x\cos^2x~dx\\&=\int t^2~dt\\&=\frac{t^3}{3}\\&=\frac{1}{3}\cos^3x\end{align*}\]
(The constant of integration can be ignored.)

The second is also a substitution, but you use an identity first:
\[\begin{align*}u_2&=\int \cos x\cos^2x~dx\\
&=\int \cos x(1-\sin^2x)~dx\\
&=\int \cos x~dx-\int\cos x\sin^2x~dx\\
&=\int\cos x~dx-\int s^2~ds&\text{where }s=\sin x\text{ and }ds=\cos x~dx\\
&=\sin x-\frac{s^3}{3}\\
&=\sin x-\frac{1}{3}\sin^3x
\end{align*}\]

Answer 9

ok i got that from using wolfram aplha. now im not sure how it equates to a solution

Answer 10

The solution itself is \(u_1y_1+u_2y_2\). It would thus be
\[y=\frac{1}{3}\cos^3x\cdot\cos x+\left(\sin x-\frac{1}{3}\sin ^3x\right)\sin x\]
or
\[y=\frac{1}{3}\cos^4x+\sin^2 x-\frac{1}{3}\sin ^4x\]
Don't forget the homogeneous solution:
\[y=\frac{1}{3}\cos^4x+\sin^2 x-\frac{1}{3}\sin ^4x+C_1\cos x+C_2\sin x\]

Answer 11

Do you see how to do the second one?

Answer 12

yes thnx ill ask if have any questions

Answer 13

no problem

Answer 14

ok im having trouble with the second one.
im at substitutions

Answer 15

Okay let me just catch up to you:
\[y''+3y'+2y=0\]
gives \[r^2+3r+2=(r+2)(r+1)=0~~\Rightarrow~~r=-2,~-1\]
So your general solutions are \(y_1=e^{-2x}\) and \(y_2=e^{-x}\).

The Wronskian:
\[\begin{vmatrix}e^{-2x}&e^{-x}\\-2e^{-2x}&-e^{-x}\end{vmatrix}=-e^{-3x}+2e^{-3x}=e^{-3x}\not=0\]

Answer 16

So, you have
\[\begin{align*}u_1&=-\int \frac{e^{-x}}{(1+e^x)e^{-3x}}~dx\\
&=-\int \frac{e^{2x}}{1+e^x}~dx
\end{align*}\]
This integral involves a somewhat tricky substitution. Let \(u=e^x\), you have \(du=e^x~dx~~\iff~~\dfrac{du}{u}=dx\). You also have \(u^2=(e^x)^2=e^{2x}\), so the integral can be rewritten as
\[\begin{align*}u_1&=-\int \frac{u^2}{1+u}\cdot\frac{du}{u}\\
&=-\int \frac{u}{1+u}~du
\end{align*}\]
Another substitution: \(z=1+u\), which gives \(u=z-1\). You have \(dz=du\), so
\[\begin{align*}u_1&=-\int \frac{u}{1+u}~du\\
&=-\int \frac{z-1}{z}~dz\\
&=-\int \left(1-\frac{1}{z}\right)~dz\\
&=\ln z-z\\
&=\ln (1+u)-(1+u)\\
&=\ln (1+e^x)-e^x-1
\end{align*}\]

Answer 17

\[\begin{align*}u_2&=\int\frac{e^{-2x}}{(1+e^x)e^{-3x}}~dx\\
&=\int\frac{e^x}{1+e^x}~dx
\end{align*}\]
This is a much simpler substitution: \(r=1+e^x\), so \(dr=e^x~dx\).
\[\begin{align*}u_2&=\int\frac{e^x}{1+e^x}~dx\\
&=\int\frac{dr}{r}\\
&=\ln r\\
&=\ln (1+e^x)
\end{align*}\]