Answer to Question #222920 in Differential Equations for Keiran

Question

xyy''+yy'+x²y'³=0

Answer 1

$given \space \\ xyy''+yy'+x^2y'^2=0( \space exact \space question \space )\\ \\ solution \space \\ this \space equation \space is \space a \space homogeneous \space polynomial \space of \space degree \space 2 \space with \space respect \space to \space y, \space y', \space y''.\\ letting \space y=e^{v(x)} \space will \space reducee \space the \space order \space of \space the \space equation \space \\ this \space gives \space \frac{dy}{dx} \space =e^{v(x)}\frac{dv}{dx} \space and \space \frac{d^2y}{dx^2}= e^{v(x)}((\frac{dv}{dx})^2+\frac{d^2v}{dx^2})\\ -----------------------------\\ e^{2v}x^2(\frac{dv}{dx})^2+e^{2v}\frac{dv}{dx}+e^{2v} \space x \space ((\frac{dv}{dx})^2+\frac{d^2v}{dx^2})=0\\ e^{2v}[x^2(\frac{dv}{dx})^2+\frac{dv}{dx}+ \space x \space ((\frac{dv}{dx})^2+\frac{d^2v}{dx^2})]=0\\ e^{2v}(x \space \frac{d^2v}{dx^2} \space + \space \frac{dv}{dx} \space +x (\space \frac{dv}{dx})^2(x+1))=0\\ e^{2v}(x^2 \space (\frac{dv}{dx})^2 \space +x \space (\frac{dv}{dx})^2 \space +x \space \space \frac{d^2v}{dx^2} \space + \space \frac{dv}{dx})=0\\ e^{2v}=0 \space and \space (x^2 \space (\frac{dv}{dx})^2 \space +x \space (\frac{dv}{dx})^2 \space +x \space \space \frac{d^2v}{dx^2} \space + \space \frac{dv}{dx})=0\\ -----------------------------\\ solve \space \\ e^{2v}=0 \\ no \space solution \space exist \space (exponential \space terms \space never \space become \space to \space zero \space so \space we \space say \space no \space solution \space exist)\\ let \space \frac{dv}{dx}=u\\ which \space gives \space \frac {d^2v}{dx^2}=\frac{du}{dx} \\ subtract \space x^2 \space u^2 \space + \space x \space u^2 \space from \space both \space sides:\\ x \space \frac {du}{dx}+u= \space -x(x+1)u^2\\ divide \space both \space sides \space by \space -xu^2\\ -\frac{\frac{du}{dx}}{u^2}-\frac{1}{xu}=x+1 \\ let \space w=\frac {1}{u} \space which \space gives \space \frac{dw}{dx}=-\frac{\frac{du}{dx}}{u^2} \\ \frac{dw}{dx}-\frac{w}{x}=x+1\\ let \space \mu= \space e^{\int \space {-\frac{1 \space }{x}dx}}=\frac{1}{x}\\ multiply \space both \space side \space by \space \mu \space \\ \frac{\frac{dw}{dx}}{x}-\frac{w}{x^2}=-(\frac{-x-1}{x})\\ substitute \space -\frac{1}{x^2}=\frac{d(\frac{1}{x})}{dx}\\ \frac{\frac{dw}{dx}}{x}+\frac{d(\frac{1}{x})}{dx}w=-(\frac{-x-1}{x})\\ apply \space the \space reverse \space product \space rule \space f \space \frac{dg}{dx} \space + \space g \space \frac{df}{dx}=\frac{d(fg)}{dx} \space to \space the \space \space left \space hand \space side \space \\ \frac{d(\frac{w}{x})}{dx}=-(\frac{-x-1}{x})\\ integrate \space both \space sides \space with \space respect \space to \space x\\ \int \space \space {\frac{d(\frac{w}{x})}{dx}}=\int \space {\frac{-(-x-1)}{x}}dx\\ evalute \space the \space integral \space \\ \frac{w}{x}=x+ \space log \space x \space + \space c_1 \space \\ divide \space both \space side \space by \space \mu \space =\frac{1}{x}\\ w=x \space (x+log \space x \space + \space c_1 \space ) \\ \space solve \space for \space u\\ u= \space \frac{1}{w}= \space \frac {1}{x(x+log \space x \space )+c_1}\\ substitute \space back \space for \space \frac{dv}{dx}=u\\ \frac{dv}{dx}=\frac{1}{x(x+log \space x)+c_1}\\ integrate \space both \space side \space with \space respect \space to \space x\\ v=\int \space {\frac{1}{x(x+log \space x)+c_1}}dx \space +c_2\\ substitute \space back \space for \space y \space = \space e^{v} \space , \space which \space give \space v= \space log \space (y)\\ log \space y=\int \space \space {\frac{1}{x(x+log \space x)+c_1}}dx \space +c_2\\ solve \space for \space y \space \\ y=e^{\int \space \space {\frac{1}{x(x+log \space x)+c_1}}dx \space +c_2}\\ it \space is \space required \space answer, \space here \space not \space neccesary \space solve \space this \space further.$