Question

find the value of b for which the given equation is exact, and then solve it using that value of b

(xy^(2)+bx^(2)y)dx+(x+y)x^(2)dy=0

Answer 1

The given equation is $(xy^{2}+bx^{2}y)dx+(x+y)x^{2}dy=0$.

Since the equation is exact we must have, $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$.

Here $M = xy^{2}+bx^{2}y; N=x^{2}(x+y)$

Then, $2xy+bx^{2} = 3x^{2}+2xy \implies b = 3$

Therefore, the given equation becomes $(xy^{2}+3x^{2}y)dx+(x+y)x^{2}dy=0$, which can be written as

$\begin{aligned} \dfrac{dy}{dx} &= \dfrac{xy^2+3x^2y}{x^2(x+y)}\\ &=\dfrac{xy(y+3x)}{x^2(x+y)}\\ &=\dfrac{y(y+3x)}{x(x+y)}\\ \end{aligned}$

This is a homogeneous equation. We use substitution to find the general solution.

Put $y=vx$. Then $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Thus,

$\begin{aligned} v+ x \frac{dv}{dx} &= \frac{vx(vx+3x)}{x(x+vx)}\\ &= \frac{v(v+3)}{(1+v)}\\ x \frac{dv}{dx} &= \frac{v(v+3)}{(1+v)} - v\\ &= \frac{v^2+3v-v-v^2}{(1+v)} \\ x \frac{dv}{dx}&= \frac{2v}{1+v} \\ \therefore \Big(\frac{1+v}{2v}\Big)dv &= \frac{dx}{x}\\ \frac{1}{2}\frac{dv}{v} + \frac{1}{2}dv &= \frac{dx}{x}\\ \frac{dv}{v}+dv &= 2 \frac{dx}{x}\\ \text{Integrating both sides,}\\ \log v + v &= 2\log x + \log c\\ \log v + v &= \log cx^2\\ v + \log\Big(\frac{v}{cx^2}\Big) &= 0\\ \log\Big(\frac{v}{cx^2}\Big) &= -v\\ \frac{v}{cx^2} &= e^{-v}\\ \frac{y}{cx^3} &= e^{-\frac{y}{x}}\\ ye^{\frac{y}{x}}&=cx^3 \end{aligned}$