Answer to Question #199523 in Differential Equations for Rajkumar

Question

Identify the following differential equations and hence solve them

i) y' = -(4/x²) - (y/x) + y²

ii) y = xy' + 1 - ln y'

Answer 1

$Question(1)\\solve\space riccati's\space equation\\ \frac{dy}{dx}=-\frac{4}{x^2}+y^2 -\frac{y}{x}\\ \\ \\ let y=-\frac{\frac{dv}{dx}}{v}, which\space gives \frac{dy}{dx}=-\frac{\frac{d^2v}{dx^2}}{v}+ \frac{(\frac{dv}{dx})^2}{v^2}:\\ -\frac{\frac{d^2v}{dx^2}}{v}+ \frac{(\frac{dv}{dx})^2}{v^2}= \frac{(\frac{dv}{dx})^2}{v^2}+\frac{\frac{dv}{dx}}{vx}-\frac{4}{x^2}\\ subtract\space \frac{(\frac{dv}{dx})^2}{v^2} from\space both \space sides\space and\space multiply\space by\space v\\ -\frac{d^2v}{dx^2}=\frac{\frac{dv}{dx}}{x}-\frac{4v}{x^2}\\ add \space \frac{d^2v}{dx^2}to \space both \space sides:\\ \frac{d^2v}{dx^2}+\frac{\frac{dv}{dx}}{x}-\frac{4v}{x^2}=0\\ multiply \space both\space sides\space by \space x^2\\ x^2\frac{d^2v}{dx^2}+x \frac{dv}{dx}-4v=0\\ assume\space a\space solution \space to\space this \space euler-cauchy \space equation\space will \space be \space proportional \space to \space x^\lambda \space for \space some \space constant \space \lambda. \\substitute \space v= x^\lambda \space into \space the \space differential\space equation \\ x^2\frac{d^2 x^\lambda}{dx^2}+x \frac{d x^\lambda}{dx}-4 x^\lambda=0\\ substitute \space \frac{d^2 x^\lambda}{dx^2}=\lambda (\lambda-1)x^{lambda-2}and\space \frac{d x^\lambda}{dx}=\lambda x^{\lambda-1}\\ \lambda^2 x^\lambda -4 x^\lambda=0\\ factor \space out x^\lambda:\\ (\lambda^2 -4 )x^\lambda=0\\ assuming\space x \ne0, the \space zeros \space must \space come \space from\space the \space polynomial:\\ (\lambda^2 -4 )=0\\ (\lambda-2)(\lambda+2)=0\\ \lambda=-2\space or\space\lambda=2 \\the \space root\space \lambda =-2 \space \\gives\space v_1=\frac{c_1}{x^2 }\space as\space a\space solotion , where \space c_1 \space is \space an\space arbitrary\space constant. \\\\the \space root\space \lambda =2 \space \\gives\space v_2=\frac{c_2}{x^2 }\space as\space a\space solution, where \space c_2 \space is \space an\space arbitrary\space constant.\\the \space general\space solution\space is \space the \space sum\space of\space the \space above\space solutions\space\\ v= v_1 +v_2 =\frac{c_1}{x}+c_2x^2\\ substitute \space back\space for\space y=-\frac{\frac{dv}{dx}}{v}\\ y=\frac{2c_1-2c_2x^4}{c_1x+c_2x^5}\\ adjust \space constant\space and\space simplify\\ y=\frac{2}{x}-\frac{4x^3}{x^4+c_1}\\ -----------------------------------------------\\Question(2)\\ solve\space the\space clairaut\space equation\space \\ y=-log(\frac{dy}{dx})+x\frac{dy}{dx}+1\\ differentiate\space both\space side\space with\space respect \space to\space x\\ \frac{dy}{dx}=x\frac{d^2y}{dx^2}+\frac{dy}{dx}-\frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}}\\ colloect\space in\space terms\space of\space \frac{d^2y}{dx^2}\\ \frac{dy}{dx}=\frac{dy}{dx}+\frac{d^2y}{dx}(x-\frac{1}{\frac{dy}{dx}})\\ subtact\space \frac{dy}{dx} \space from \space both\space sides\\ \frac{d^2y}{dx}(x-\frac{1}{\frac{dy}{dx}})=0\\ solve\space \frac{d^2y}{dx}=0 \space and \space x-\frac{1}{\frac{dy}{dx}}=0 \space separately\\ for \space \frac{d^2y}{dx}=0\\ integrate \space both\space sides\space with\space wrt \space x\\ \frac{dy}{dx}=\int 0\space dx= c_1 , where \space c_1 is\space an\space arbitrary\space constant.\\ subsitute \frac{dy}{dx}=c_1 into\space y=-log\frac{dy}{dx}+x\frac{dy}{dx}+1\\ y=-log(c_1)+c_1 x+1\\ for \space x-\frac{1}{\frac{dy}{dx}}=0\\ solve \space for\space \frac{dy}{dx}\\ \frac{dy}{dx}=\frac{1}{x} \\ substitute \space into\space y=-log\frac{dy}{dx}+x\frac{dy}{dx}+1\\ y=-log(\frac{1}{x})+2\\ y=-log(\frac{1}{x})+2 \space or y=-log(c_1)+c_1x+1$

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