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

Marks : 6

Year : 2015

$$y^4 dx=(x^{-3/4}-y^3x)dy$$ $$ \dfrac {dx}{dy}=\dfrac {x^{-3/4}}{y^4}-\dfrac xy$$

Multiplying $x^{3/4}$ Throughout

$ \therefore x^{3/4}\dfrac {dx}{dy}=\dfrac 1{y^4}-\dfrac {x^{7/4}}y\\ \therefore x^{3/4}\dfrac {dx}{dy}-\dfrac {x^{7/4}}y=\dfrac 1{y^4}------ (1)\\ Let \space v=x^{7/4}\\ \dfrac {dv}{dy}=\dfrac 74x^{3/4}\dfrac {dx}{dy}\\ \therefore \dfrac 47 \dfrac {dv}{dy}-\dfrac vy=\dfrac 1{y^4}\\ \dfrac {dv}{dy}-\dfrac 74\dfrac vy=\dfrac 74\dfrac 1{y^4}\\ $

Which is similar to

$\dfrac {dv}{dy}-pv=Q \\ where \space P=\dfrac 74\dfrac 1y=\dfrac 7{4y} \space \& \space Q=\dfrac 7{4y^4}\\ I.F =e^{\int pdy}\\ =e^{\int \dfrac 7{4y}dy}\\ =e^{\dfrac 74\log y}\\ =e^{\log y 7/4} \\ = y^{\frac 74}\\ \text{Solution is}\\ v.I.F=\int Q.IF \space dy \\ v.y^{7/4}=\int \dfrac 7{4y^4}y^{7/4}dy\\ =\dfrac 74\int y^{-9/4}dy\\ =\dfrac 74\dfrac {y^{-5/4}}{\dfrac {-5}4}+c\\ =\dfrac{-7}5\dfrac 1{y^{5/4}}+c $

Resubstituting value of V

$x^{7/4}y^{7/4}+\dfrac 7{5y^{5/4}}=c$