Originally Posted by

**aspotonthewall** I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it.

So I'll write out, step by step, what I've done:

$\displaystyle y''-21-7x^2+3y^2+x^2y^2=0$

$\displaystyle => y''=(7-y^2)(x^2+3)$

$\displaystyle =>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx}$

Right Hand Side:

$\displaystyle \int{x^2+3 dx}=\frac{x^3}{3}+3x+c$

Left Hand Side:

Rewriting LHS using partial fractions:

$\displaystyle \frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7 }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)}$$\displaystyle =\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}$

$\displaystyle A(\sqrt{7}+y)+B(\sqrt{7}-y)=1$

Let $\displaystyle y=\sqrt{7}$

$\displaystyle A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1$

$\displaystyle => A(2\sqrt{7})=1$

$\displaystyle => A=\frac{1}{2\sqrt{7}}$

Let $\displaystyle y=-\sqrt{7}$

$\displaystyle A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1$

$\displaystyle =>B(2\sqrt{7})=1$

$\displaystyle => B=\frac{1}{2\sqrt{7}}$

So then I get LHS to be:

$\displaystyle \frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y})$

Integrating the LHS with respect to y gives

$\displaystyle \frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c$

So now I have

$\displaystyle \frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt {7}-y)})}=\frac{x^3}{3}+3x$

$\displaystyle =>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c$

But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong.