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

Marks : 6

Year : 2014

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

Dividing by $x^2$ through out we get

$$\dfrac 1{x^2}\dfrac {dx}{dy} - \dfrac 1x y=y^3 -------(1) \ \text {Let} \dfrac {-1}x = v \ $$

$\therefore$ Differentiating w.r.t y

we get

$\dfrac 1{x^2}\dfrac {dx}{dy}=\dfrac {dv}{dy}$

substituting these values in eq (1)

we get

$$\dfrac {dv}{dy}+vy=y^3$$ which is same as that of $$\dfrac {dv}{dy}+P_y=Q\ \text{where}\space P=Y \space \text{and }\space Q=Y^3\ I.F=e^{\int p\space dy} \= e^{\int p\space dy}\ = e^{\frac {y^2}2}$$

$\therefore $ Solution is

$$V\times I.F=\int Q.I.F. dy\ V.e^{\frac {y^2}2}=\int y^3e^{\frac {y^2}2}dy\ Let \dfrac {y^2}2=U\ \therefore y^{dy}=du\ Ve^{\frac {y^2}2}=\int 2Ue^U\space du\ =2[Ue^U-e^u]+ c\ =2e^U(U-1)+c$$ Resubstituting value of U $$V.e^{\frac {y^2}2}=2e^{\frac {y^2}2}\Bigg(\dfrac {y^2}2-1\Bigg)+C\ \dfrac {-1}xe^{\frac {y^2}2}=2e^{\frac {y^2}2}\Bigg(\dfrac {y^2}2-1\Bigg) + C \ \therefore \dfrac {-1}x=2\Bigg(\dfrac {y^2}2-1\Bigg)+ ce^{\frac {-y^2}2}\ \therefore -1=2x\Bigg(\dfrac {y^2}2-1\Bigg) =cxe^{\frac {-y^2}2}\ x(y^2-2)+1=cxe^{\frac {-y^2}2}$$