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Question Number 34724 by mondodotto@gmail.com last updated on 10/May/18

$$\mathbf{use}\mathbf{bernoulli}\mathbf{methods}$$$${\mathbf{sec}}^2\mathit{y}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+\mathit{x}\mathbf{tan}\mathit{y}={\mathit{x}}^3$$

Commented by mondodotto@gmail.com last updated on 10/May/18

$$\mathrm{please}\mathrm{someone}\mathrm{solve}\mathrm{this}$$

Answered by candre last updated on 10/May/18

$$v=\tan y$$$$\frac{dv}{dx}={\sec }^{2}y\frac{dy}{dx}$$$$\frac{dv}{dx}+xv=x^3$$$$\mu \left(x\right)=e^{\int xdx}=e^{x^2/2}$$$$e^{x^2/2}{v}^{\prime }+{xe}^{x^2/2}v=e^{x^2/2}{x}^3$$$$\int {\left(e^{x^2/2}v\right)}^{\prime }dx=\int x^3{e}^{x^2/2}dx$$$$e^{x^2/2}v=e^{x^2/2}(x^2-2)+C$$$$v=x^2-2+{Ce}^{-x^2/2}$$$$\tan y=x^2-2+{Ce}^{-x^2/2}$$

Commented by candre last updated on 10/May/18

$${\sec }^{2}y\frac{dy}{dx}=2x-{Cxe}^{-x^2/2}$$$$x\tan y=x^3-2x+{Cxe}^{-x^2/2}$$$${\sec }^{2}y\frac{dy}{dx}+x\tan y=x^3$$

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