Answer to Question #222285 in Differential Equations for mercy

Question

prove that the substitution u=y^1-n reduces the equation dy/dx+p(x)y=Q(x)yⁿ into a linear equation . hence solve the equation dy/dx+y/x=x/y³

Answer 1

Part 1: proving$Divide\ the\ equation\ \frac{dy}{dx}+P(x)y=Q(x)y^{n}\ by\ y^{n}\newline \frac{1}{y^{n}}\frac{dy}{dx}+P(x)y^{1-n}=Q(x)\newline set\ u=y^{1-n},\ \frac{du}{dx}=(1-n)y^{1-n-1}\frac{dy}{dx}\newline \frac{1}{1-n}\frac{du}{dx}=\frac{1}{y^{n}}\frac{dy}{dx}\newline substitute\ to\ get\ \frac{1}{1-n}\frac{du}{dx}+P(x)u=Q(x)\newline \frac{du}{dx}+(1-n)P(x)u=(1-n)Q(x)\newline which\ is\ linear\ in\ x\newline Part 2\newline \frac{dy}{dx}+P(x)y=Q(x)y^{n}\newline \frac{dy}{dx}+\frac{y}{x}=\frac{x}{y^{3}}\newline n=-3,\ P(x)=\frac{1}{x},\ Q(x)=x\newline \frac{du}{dx}+(1-n)uP(x)=(1-n)Q(x)\newline \frac{du}{dx}+\frac{4u}{x}=4x\newline let\ u=vw\newline \frac{du}{dx}=V\frac{dw}{dx}+W\frac{dv}{dx}\newline substitute\newline V\frac{dw}{dx}+W\frac{dv}{dx}+4\frac{VW}{x}=4x\newline V\frac{dw}{dx}+W(\frac{dv}{dx}+\frac{4v}{x})=4x\newline take\ part\ in\ ()\ equals\ 0\ and\ separate\ variables\newline \frac{dv}{dx}+\frac{4v}{x}=0\newline \frac{dv}{v}=\frac{-4dx}{x}\newline \int \frac{dv}{v}=-\int \frac{4dx}{x}\newline ln (v)= ln(k)-4ln(x)\newline v=kx^{-4}\newline substitute\ v\ back\newline kx^{-4}\frac{dw}{dx}=4x\newline kx^{-4}dw=4xdx\newline kdw=4x^{9}dx\newline \int kdw=\int 4x^{9}dx\newline kw=\frac{4}{10}x^{10}+C\newline w=\frac{1}{k}(\frac{4}{10}x^{10}+C)\newline substitute\ u=vw\ to\ find\ original\ equation\newline u=VW=\frac{kx^{-4}}{k}(\frac{4}{10}x^{10}+C)\newline =x^{-4}(\frac{4}{10}x^{10}+C)\newline \frac{4}{10}x^{6}+Cx^{-4}\newline u=y^{1-n}=y^{4}\newline We\ then\ substitute\ back\ y=u^{\frac{1}{4}}\newline y=(\frac{4}{10}x^{6}+Cx^{-4})^{\frac{1}{4}}$