Browse Questions

# Solve the following differential equation : $xdy-(y+2x^2)dx=0$

Toolbox:
• If the equation is of the form $\large\frac{dy}{dx}$$+ py= Q, then the integrating factor is e^{\large\int pdx} Step 1: xdy -(y+2x^2)dx=0 xdy=(y+2x^2)dx x\large\frac{dy}{dx}=$$y+2x^2$
Step 2:
Divide throughout by $x$
$\large\frac{dy}{dx}=\frac{y}{x}$$+2x \large\frac{dy}{dx}-\frac{y}{x}$$=2x$
Clearly this is a homogeneous linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q The integrating factor I.F is e^{\int Pdx}=e^{\int -\large\frac{1}{x}dx} e^{-\log x}=\large\frac{1}{x} Step 3: The required solution is ye^{\int Pdx}=\int Qe^{\int Pdx}.dx+c (i.e) y\big(\large\frac{1}{x}\big)=$$\int 2x\big(\large\frac{1}{x}\big)$$dx+c \Rightarrow \large\frac{y}{x}=$$2x+c$
$y-2x^2+cx=0$ is the required solution.