#### tapout1829

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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:

- Thread starter tapout1829
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- Mar 20, 2006

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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:

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.

Hello, tapout1829!

Are you in self-study?

This should have been explained in your class.

This problem is similar to one you had in Algebra I.Determine whether \(\displaystyle y\:=\:x^2e^x\) is a solution of the differential equation: \(\displaystyle \,xy'\,-\,2y\:=\:x^3e^x\)

\(\displaystyle \;\;\)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!