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 with Step-by-step explanation:We are given that [tex](21x^2y+2xy+7y^3)dx+(x^2+y^2)dy=0[/tex]Compare with [tex]Mdx+Ndy=0[/tex]Then, we get [tex]M=21x^2y+2xy+7y^3[/tex][tex]N=x^2+y^2[/tex]Differentiate M w.r.t y[tex]M_y=21x^2+2x+21y^2[/tex]Differentiate N w.r.t x[tex]N_x=2x[/tex][tex]\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[/tex]Q=21[tex]I.F=e^{\int 21 dx}=e^{21x}[/tex][tex]M=e^{21x}(21x^2+2xy+7y^3)[/tex][tex]N=e^{21x}(x^2+y^20[/tex]Solution is given by [tex]\int_{y\;constant} Mdx+\int_{x\;free\;terms} Ndy=C[/tex][tex]\int_{y\;constant} e^{21x}(21x^2y+2xy+7y^3)dx=C[/tex][tex]\int_{y\;constant}21x^2ye^{21x} dx+\int_{y\;constant}2xye^{21x}dx+\int_{y\;constant}7y^3e^{21x}dx=C[/tex]Using partial integration [tex]u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx[/tex][tex]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[/tex][tex]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[/tex][tex]x^2ye^{21x}+\frac{y^3e^{21x}}{3}=C[/tex]