Originally Posted by

**r_maths** I'm doing something wrong, could someone point me to the right direction...

Find the general solution of the differential equation.

$\displaystyle x\frac{dy}{dx}-3y=x^{4}$

**ANSWER: **$\displaystyle y=x^{3}(x+c)$

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$\displaystyle x\frac{dy}{dx}-3y=x^{4}$

$\displaystyle -3y=x^{4}-x\frac{dy}{dx}$

$\displaystyle 3y=x\frac{dy}{dx}-x^{4}$

$\displaystyle \frac{3y}{dy}=\frac{x}{dx}-x^{4}$ SiMoon says : why didn't you divide $\displaystyle x^4$ by dy ?

$\displaystyle \int{\frac{3y}{dy}}=\int{\frac{x}{dx}}-\int{x^{4}}$

SiMoon says : these two last integrals make no sense, you don't know how to integrate when dx or dy is in the denominator

$\displaystyle \frac{3}{2}y^{2}=\frac{1}{2}x^{2}-\frac{1}{5}x^{5}+c$

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