Question

If My−NxN=Q, where Q is a function of x only, then the differential equation M+Ny′=0 has an integrating factor of the form μ(x)=e∫Q(x)dx. Find an integrating factor and solve the given equation. (21x2y+2xy+7y3)dx+(x2+y2)dy=0

Answer 1

Answer with Step-by-step explanation:

We are given that

$(21x^2y+2xy+7y^3)dx+(x^2+y^2)dy=0$

Compare with

$Mdx+Ndy=0$

Then, we get

$M=21x^2y+2xy+7y^3$

$N=x^2+y^2$

Differentiate M w.r.t y

$M_y=21x^2+2x+21y^2$

Differentiate N w.r.t x

$N_x=2x$

$\frac{M_y-N_x}{N}=\frac{21x^2+2x+21y^2-2x}{x^2+y^2}=\frac{21(x^2+y^2)}{x^2+y^2}=21$

Q=21

$I.F=e^{\int 21 dx}=e^{21x}$

$M=e^{21x}(21x^2+2xy+7y^3)$

$N=e^{21x}(x^2+y^20$

Solution is given by

$\int_{y\;constant} Mdx+\int_{x\;free\;terms} Ndy=C$

$\int_{y\;constant} e^{21x}(21x^2y+2xy+7y^3)dx=C$

$\int_{y\;constant}21x^2ye^{21x} dx+\int_{y\;constant}2xye^{21x}dx+\int_{y\;constant}7y^3e^{21x}dx=C$

Using partial integration

$u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

$21x^2y}\frac{e^{21x}}{21}-2y\int xe^{21x}dx+\frac{2xye^{21x}}{21}-\frac{2y}{21}\int e^{21x}dx+\frac{7y^3e^{21x}}{21}=C$

$x^2ye^{21x}-\frac{2xye^{21x}}{21}+\frac{2ye^{21x}}{441}+\frac{2xye^{21x}}{21}-\frac{2ye^{21x}}{441}+\frac{7y^3e^{21x}}{21}=C$

$x^2ye^{21x}+\frac{y^3e^{21x}}{3}=C$