Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2013

$$\dfrac {x+2y^3}y=\dfrac {dx}{dy}$$

$\therefore \dfrac xy+2y^2=\dfrac {dx}{dy} \\ \therefore \dfrac {dx}{dy}-\dfrac xy=2y^2 \\ \text{Which is similar to } \dfrac {dx}{dy}+Px=Q \\ \text{Where} P=P=\dfrac {-1}y \space \&\space Q=2y^2 \\ I.F=e^{\int Pdy} \\ =e^{\int \dfrac {-1}y dy} \\ =e^{-\log y} \\ =e^{\log\frac 1y} \\ =\dfrac 1y \\ \text{Solution is } \\ x.I.F=\int Q.IF.dy \\ x.\dfrac 1y=\int 2y^2\times \dfrac 1ydy \\ \dfrac xy=\int 2y\space dy \\ \dfrac xy=y^2+C \\ \therefore x=y^3+cy $