# How to find the solution of the differential equation

Find the solution of the differential equation $$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$

Trial: \begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ \implies v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)} \end{align} I can't seperate $x$ and $v$.

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Follow the procedure here. –  David Mitra Jan 23 '13 at 15:49

## 2 Answers

Hint: It is an exact equation. Assume there is an differentiable function $f(x,y)$ such that $$f_x=x^3+xy^2-10x,~~~f_y=x^2y+y^3+5y$$ and then find the function. The solution is as $$f(x,y)=C$$

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Nice job..."jack of all trades!" +1 –  amWhy Feb 8 '13 at 5:24

Your differential equation can be written as $$(x^{3}+x y^{2}-10 x)dx+(x^{2}y+y^{3}+5y)dy=0$$ So it is of the form $M(x,y)dx+N(x,y)dy=0$ Now, $$M_{y}=2xy$$ and $$N_{x}=2xy.$$ $\therefore$ The given differential equation is exact. So its general solution is $$\int M dx+\int (\text{Terms in N which does not contain x})dy=C$$ $$\Rightarrow \int (x^{3}+x y^{2}-10 x)dx+\int 0 dy=C$$ $$\Rightarrow \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=c$$ Now $y(0)=1$, we have $C=0$.

$\therefore$ The solution of given differential equation is $\displaystyle \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=0$.

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I don't understand this part $\int (\text{Terms in$N$which does not contain$x$})dy$. By the way answer is $2x^2y^2+x^4+y^4+10y^2-20x^2-11=0$ –  Argha Jan 26 '13 at 18:59