First Order Linear Differential Equations

[wubie]

Hello,

It has been over a year since I last did calculus. And I am having trouble with my current calculus course. First here is the question:

Solve the given first-order linear equation and verify that your solution indeed satisfies the equation.

y' - 2xy = 2xe^x^2


Now I THINK I have the answer:

y = e^x^2 ( x^2 + c)

But how do I verify? I would think I simply would take the above equation and it's derivative and sub. into the equation

y' - 2xy = 2xe^x^2

If that is the case, my problem is this: When I take the derivative of

y = e^x^2 ( x^2 + c)

I have

y' = 2x * e^x^2 + 2x^3 * e^x^2 + 2x * C * e^x^2

How can I verify my answer when there is an unknown constant in my derivative? What am I missing?

Any help is appreciated - detailed if possible. Thankyou.

 

[jamesrc]

Keep doing what you're doing and the constant will cancel out:

y = e^{x^2}(x^2+{\rm C})
y' = 2x{\rm C}e^{x^2} + 2x^3e^{x^2} + 2xe^{x^2}
2xy = 2x{\rm C}e^{x^2} + 2x^3e^{x^2}

Substitute into find that:

y'-2xy = 2xe^{x^2}

 

[Kurdt]

Ok, as I understand it we have the following equation that we must solve.

\frac {dy}{dx}-2xy=2xe^{x^2}

First we must compute the integrating factor which is.

F(x)=e^{\int{f(x)dx}}

and multiply the original equation by this to get

e^{x^2}\frac{dy}{dx}-2xye^{x^2}=2xe^{2x^2}

And then the final equation to solve just becomes

ye^{x^2}=\int2xe^{2x^2}dx

and y is then

y=-\frac{e^{2x^2}}{2}+C

 

[wubie]

Thanks to the both of you.

I think kurdt that you made an error.

f(x) = - 2x

not

f(x) = 2x

Thanks anyway though.

Cheers.

 

[Kurdt]

Oops I apologise. Well it just goes to prove I am only human