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Question Number 63894 by mathmax by abdo last updated on 10/Jul/19

$$sovethe\left(de\right)x^2{y}^{{}^{\prime }}-(2x+3)y=sin\left(x^2\right)withy\left(1\right)=2and$$$$y^{{}^{\prime }}\left(1\right)=1.$$

Commented by mathmax by abdo last updated on 11/Jul/19

$$\left(he\right)\to x^2{y}^{{}^{\prime }}-(2x+3)y=0\Rightarrow x^2{y}^{{}^{\prime }}=(2x+3)y\Rightarrow$$$$\frac{y^{{}^{\prime }}}{y}=\frac{2x+3}{x^2}=\frac{2}{x}+\frac{3}{x^2}\Rightarrow ln\mid y\mid =2ln\mid x\mid -\frac{3}{x}+c\Rightarrow$$$$y=K{x}^2{e}^{-\frac{3}{x}}letusemvcmethod$$$$y^{{}^{\prime }}=K^{{}^{\prime }}{x}^2{e}^{-\frac{3}{x}}+K\{2x{e}^{-\frac{3}{x}}+x^2\frac{3}{x^2}{e}^{-\frac{3}{x}}\}$$$$=K^{{}^{\prime }}{x}^2{e}^{-\frac{3}{x}}+K\{2x+3\}{e}^{-\frac{3}{x}}$$$$\left(e\right)\Rightarrow K^{{}^{\prime }}{x}^4{e}^{-\frac{3}{x}}+{Kx}^2(2x+3)e^{-\frac{3}{x}}-(2x+3){Kx}^2{e}^{-\frac{3}{x}}=sin\left(x^2\right)\Rightarrow$$$$K^{{}^{\prime }}{x}^4{e}^{-\frac{3}{x}}=sin\left(x^2\right)\Rightarrow K^{{}^{\prime }}=\frac{sin\left(x^2\right)e^{\frac{3}{x}}}{x^4}\Rightarrow$$$$K\left(x\right)={\int }_1^x\frac{sin\left(t^2\right)e^{\frac{3}{t}}}{t^4}dt+\lambda$$$$\lambda =K\left(1\right)wehavey\left(1\right)=K\left(1\right)e^{-3}\Rightarrow K\left(1\right)=e^{3}y\left(1\right)\Rightarrow$$$$K\left(x\right)={\int }_1^x\frac{sin\left(t^2\right)e^{\frac{3}{t}}}{t^4}dt+2e^3\Rightarrow$$$$y\left(x\right)=x^2{e}^{-\frac{3}{x}}\{{\int }_1^x\frac{e^{\frac{3}{t}}sin\left(t^2\right)}{t^4}dt+2e^3\}....$$

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