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Question Number 112456 by bobhans last updated on 08/Sep/20

$$\mathrm{solve}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{4\mathrm{y}}{\mathrm{x}}=1+\frac{2}{\mathrm{x}}$$

Answered by john santu last updated on 08/Sep/20

$$letu={yx}^{-4}\to \frac{du}{dx}=-4{yx}^{-5}+x^{-4}\frac{dy}{dx}$$$$\frac{dy}{dx}=x^4\frac{du}{dx}+4{yx}^{-1}$$$$\iff x^4\frac{du}{dx}+\frac{4y}{x}-\frac{4y}{x}=1+\frac{2}{x}$$$$\iff \frac{du}{dx}=\frac{1}{x^4}+\frac{2}{x^5};du=(x^{-4}+2x^{-5})dx$$$$\int du=\int (x^{-4}+2x^{-5})dx$$$$\Rightarrow u=-\frac{1}{3x^3}-\frac{1}{2x^4}+c$$$$\Rightarrow \frac{y}{x^4}=-\frac{1}{3x^3}-\frac{1}{2x^4}+C$$$$\Rightarrow y=-\frac{1}{3}x-\frac{1}{2}+{Cx}^4$$

Answered by abdomsup last updated on 08/Sep/20

$$h\to y^{{}^{\prime }}-\frac{4}{x}y=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{4}{x}\Rightarrow$$$$ln\mid y\mid =4ln\mid x\mid +c\Rightarrow y=k{x}^4$$$$lagrangemethod{y}^{{}^{\prime }}=k^{{}^{\prime }}{x}^4+4{kx}^3$$$$e\Rightarrow k^{{}^{\prime }}{x}^4+4k{x}^3-4{kx}^3=1+\frac{2}{x}\Rightarrow$$$$k^{{}^{\prime }}=x^{-4}+2x^{-5}\Rightarrow$$$$k=-\frac{1}{3}{x}^{-3}-\frac{1}{2}{x}^{-4}+\lambda \Rightarrow$$$$y=(-\frac{1}{3}{x}^{-3}-\frac{1}{2}{x}^{-4}+\lambda )x^4$$$$=-\frac{x}{3}-\frac{1}{2}+\lambda x^4$$$$$$

Answered by Dwaipayan Shikari last updated on 08/Sep/20

$$I.F=e^{\int \frac{-4}{x}}=\frac{1}{x^4}$$$$\frac{y}{x^4}=\int \frac{1}{x^4}+\frac{2}{x^5}$$$$y=-\frac{x}{3}-\frac{1}{2}+{Cx}^4$$

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