Solve 2.1.5 DE w/ $y(0)=1$ & Find $\mu(x)$

[karush]

find y
$$\displaystyle y^\prime - y =2xe^{2x}, \quad y(0)=1$$

obtain u(x)
$$\displaystyle\mu(x)
=\exp\left(\int (1),dx\right)
=e^{x}$$

if ok then proceed to distribute to each term
also, this has $y(0)=1$ so new feature to deal with?

 

[MarkFL]

Careful...in this case $P(x)=-1$, and so:

\(\displaystyle \mu(x)=\exp\left(-\int\,dx\right)=e^{-x}\)

:)

edit: Being given the initial value $y(0)=1$ turns this into an IVP, which will allow you to determine the constant of integration to get a specific solution that both satisfies the ODE and the given initial condition.

 

[karush]

Careful...in this case $P(x)=-1$, and so:

\(\displaystyle \mu(x)=\exp\left(-\int\,dx\right)=e^{-x}\)

:)

edit: Being given the initial value $y(0)=1$ turns this into an IVP, which will allow you to determine the constant of integration to get a specific solution that both satisfies the ODE and the given initial condition.

distribute $e^{-x}$
$$\displaystyle e^{-x}y^\prime - e^{-x}y =2xe^{2x}e^{-x}=2e^x x$$
$$\frac{dy}{dx}(xy)=2e^x x$$
integrate
$$xy=\int 2e^x x \, dx =2e^x(x-1)+c $$
divide by x
$$y=2e^x-\frac{2e^x}{x}+\frac{c}{x}$$

 

[MarkFL]

distribute $e^{-x}$
$$\displaystyle e^{-x}y^\prime - e^{-x}y =2xe^{2x}e^{-x}=2e^x x$$
$$\frac{dy}{dx}(xy)=2e^x x$$
integrate
$$xy=\int 2e^x x \, dx =2e^x(x-1)+c $$
divide by x
$$y=2e^x-\frac{2e^x}{x}+\frac{c}{x}$$

What you want is:

\(\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=2xe^x\)

Now you can integrate.

Yous see, after multiplying through by the integrating factor, you should have:

\(\displaystyle \frac{d}{dx}\left(\mu(x)y\right)=\mu(x)Q(x)\)

 

[karush]

$$\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=2xe^x$$
$\textit{integrate } $
$$\displaystyle e^{-x}y=\int 2xe^x dx$$
$$\displaystyle e^{-x}y =2e^x(x-1)+c$$
$\textit{divide by $e^{-x}$}$
$$\displaystyle y=2e^{2x}(x-1)+\frac{c}{e^{-x}}$$
$\textit{so if $y(0)=1$ then}$something goofy!

 

[MarkFL]

I would write the general solution to the ODE as:

\(\displaystyle y(x)=2(x-1)e^{2x}+c_1e^x\)

Now, we can write:

\(\displaystyle y(0)=2(0-1)e^{2\cdot0}+c_1e^0=1\)

Solve this for $c_1$...what do you get?

 

[karush]

\(\displaystyle c_1= 0\)

$\textit{so does this mean $c_1$ can be dropped}$

 

[MarkFL]

\(\displaystyle c_1= 0\)

$\textit{so does this mean $c_1$ can be dropped}$

Please check your computation...I get a different result. :)