Diff equation (xy' - 2y = x^3 e^x)

[tapout1829]

Hey guys i need help to do this problem:

Determine wheather the function is a solution of the differential equation --> xy' - 2y = x^3e^x

y = x^2e^x is the question

thanks guys :roll:

 

[galactus]

To see if $\displaystyle y=x^{2}e^{x}$ is a solution, take it and its derivative

and sub into the left side of your DE. If you get the right side, then it's a

solution.


If you need to know how to attempt deriving the answer, then first find your integrating factor.

\(\displaystyle \L\\x\frac{dy}{dx}-2y=x^{3}e^{x}\)

Divide through by x:

\(\displaystyle \L\\\frac{dy}{dx}-\frac{2}{x}y=x^{2}e^{x}\)

Now, can you find the integrating factor and finish?. It ain't too bad.

 

[soroban]

Re: Diff equation

Hello, tapout1829!

Are you in self-study?
This should have been explained in your class.

Determine whether $\displaystyle y\:=\:x^2e^x$ is a solution of the differential equation: $\displaystyle \,xy'\,-\,2y\:=\:x^3e^x$

This problem is similar to one you had in Algebra I.
$\displaystyle \;\;$Determine whether $\displaystyle x\,=\,2$ is a solution of: $\displaystyle \,2x^2\,-\,x\:=\:6$

We "plug in" $\displaystyle x\,=\,2$ and see if we get a true statement, right?
We replace every $\displaystyle x$ with $\displaystyle 2$ and see what we get.

The left side is: $\displaystyle \,2\cdot2^2\,-\,2$ . . . Does this equal the right side?
We have: $\displaystyle \,8\,-\,2\:=\:6$ . . . yes!

They already did the hard work (solving for $\displaystyle x$); we just had to verify the answer.


This is problem is the same.
They already solved the differential equation,
$\displaystyle \;\;$and they want us to verify their answer.

We have: \(\displaystyle \,xy'\,-\2y\:=\:x^3e^x\)
Answer: $\displaystyle y\:=\:x^2e^x$

"Plug in" the answer and see if we get a true statement.
Replace $\displaystyle y$ with $\displaystyle x^2e^x$ . . . Replace $\displaystyle y'$ with $\displaystyle x^2e^x\,+\,2xe^x$

The left side is: \(\displaystyle \,x\left(x^2e^x\,+\,2xe^x)\,-\,2(x^2e^x)\) . . . Does this equal the right side?

We have: $\displaystyle \,x^3e^x\,+\,2x^2e^x \,-\,2x^2e^x\;=\;x^3e^x$ . . . yes!