# Thread: Help finding solution to first-order differential equation

1. ## Help finding solution to first-order differential equation

I have the differential equation:

$(x^4+y^2)dx-xydy = 0$

and I am really struggling to solve it. I have attempted to use the integrating factor method to make it an exact equation but either through algebraic error or it being the incorrect method of finding a solution I have come unstuck.

Any help would be much appreciated.

Thanks

2. ## Re: Help finding solution to first-order differential equation

This one?
Originally Posted by zengodspeed
$$x^4+y^2 \,\mathrm dx-xy \, \mathrm dy=0$$

3. ## Re: Help finding solution to first-order differential equation

Do you mean $\displaystyle (x^4+ y^2)dx- xy dy= 0$?

4. ## Re: Help finding solution to first-order differential equation

\begin{align*} (x^4+ y^2) \, \mathrm dx- xy \, \mathrm dy &= 0 \implies \left\{\begin{aligned}M(x,y) &= x^4 + y^2 \\ N(x,y) &= -xy \end{aligned}\right. \\[8pt] h(x) &= \frac{M_y - N_x}{N} \\ &= \frac{2y - (-y)}{-xy} \\ &= -\frac{3y}{xy} \\ &= -\frac3x \\[8pt] u(x) &= e^{\int h(x) \, \mathrm dx} \\ &= e^{-3\ln x} \\ &= x^{-3} \\[8pt] (x+x^{-3}y^2 ) \, \mathrm dx- x^{-2}y \, \mathrm dy &= 0 \implies \left\{\begin{aligned}M(x,y) &= x+x^{-3}y^2 \\ N(x,y) &= - x^{-2}y \end{aligned}\right. \\ M_y &= 2x^{-3}y = N_x \implies \text{exact} \\[8pt]
F(x,y) &= \int N(x,y) \,\mathrm dy \\ &= \int -x^{-2}y \,\mathrm dy \\ &= -\tfrac12x^{-2}y^2 + f(x) \\ F_x &= x^{-3}y^2 + f'(x) = x + x^{-3}y^2 = M(x,y) \\ f'(x) &= x \\ f(x) &= \tfrac12x^2 \\ F(x,y) = -\tfrac12x^{-2}y^2 + \tfrac12x^2 &= c_1 \\ x^{-2}y^2 + x^2 &= c &(c = 2c_1) \\ y^2 &= cx^2 - x^4 \\ &= (c-x^2)x^2 \\ y &= \pm x\sqrt{c-x^2}\end{align*}