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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

Please find the general solution of :

$xy'-2y =x^2$

$xy'-2y =x^2$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

Please find the general solution of :

$xy'-2y =x^2$

$xy'-2y =x^2$

- Admin
- #2

\(\displaystyle x^{-2}y'-2x^{-3}y=x^{-1}\)

Now we may observe that the left have side is the derivative of a product:

\(\displaystyle \frac{d}{dx}\left(x^{-2}y \right)=x^{-1}\)

Integrate with respect to $x$:

\(\displaystyle \int\frac{d}{dx}\left(x^{-2}y \right)\,dx=\int x^{-1}\,dx\)

\(\displaystyle x^{-2}y=\ln|x|+C\)

Thus, we find the general solution is:

\(\displaystyle y(x)=x^2\left(\ln|x|+C \right)\)

\(\displaystyle \displaystyle \begin{align*} x\,\frac{dy}{dx} - 2y &= x^2 \\ \frac{dy}{dx} - \frac{2}{x}\,y &= x \end{align*}\)

which is now a first order linear DE. The integrating factor is

\(\displaystyle \displaystyle \begin{align*} e^{ \int{ -\frac{2}{x} \, dx } } = e^{ -2\ln{(x)} } = e^{ \ln{ \left( x^{-2} \right) } } = x^{-2} \end{align*} \)

so multiplying both sides of our linear DE by the integrating factor gives

\(\displaystyle \displaystyle \begin{align*} x^{-2}\,\frac{dy}{dx} - 2x^{-3}\,y &= \frac{1}{x} \end{align*}\)

which is the same as multiplying the original equation by \(\displaystyle \displaystyle \begin{align*} x^{-3} \end{align*}\).

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- #4

- Jan 26, 2012

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