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`

`(M_y-N_x)/(N)=(21x^2+2x+21y^2-2x)/(x^2+y^2)=(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 ((du)/(dx)\int vdx)dx`

`21x^2y}(e^(21x))/(21)-2y\int xe^(21x)dx+(2xye^(21x))/(21)-(2y)/(21)\int e^(21x)dx+(7y^3e^(21x))/(21)=C`

`x^2ye^(21x)-(2xye^(21x))/(21)+(2ye^(21x))/(441)+(2xye^(21x))/(21)-(2ye^(21x))/(441)+(7y^3e^(21x))/(21)=C`

`x^2ye^(21x)+(y^3e^(21x))/(3)=C`