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Question Number 84232 by niroj last updated on 10/Mar/20

$$\mathbf{Solve}\mathbf{the}\mathbf{following}\mathbf{differential}\mathbf{equation}.$$$$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}+\frac{\mathbf{x}}{1-{\mathbf{x}}^2}\frac{\mathbf{dy}}{\mathbf{dx}}-\frac{\mathbf{y}}{1-{\mathbf{x}}^2}=\mathbf{x}\sqrt{1-{\mathbf{x}}^2}$$

Commented by mind is power last updated on 11/Mar/20

$$checksirpleaseifisnot$$$$\frac{d^{2}y}{{dx}^2}-\frac{x}{1-x^2}.\frac{dy}{dx}-\frac{y}{1+x^2}=x\sqrt{1-x^2}?$$

Commented by niroj last updated on 11/Mar/20

$$trytocomplete.sir.$$

Answered by mind is power last updated on 11/Mar/20

$$if$$$$\frac{d^{2}y}{{dx}^2}-\frac{x}{1-x^2}.\frac{dy}{dx}-\frac{y}{1-x^2}=x\sqrt{1-x^2}$$$$x=sin\left(t\right)\Rightarrow y\left(x\right)=y\left(sin\left(t\right)\right)=z\left(t\right)$$$$\frac{dy}{dx}=\frac{d}{dt}y\left(sin\left(t\right)\right).\frac{1}{\frac{dx}{dt}}=z^{\prime }\left(t\right).\frac{1}{cos\left(t\right)}$$$$\frac{d^{2}y}{{dx}^2}=\frac{z^{″}cos\left(t\right)+sin\left(t\right)z^{\prime }}{{cos}^2\left(t\right)}.\frac{1}{cos\left(t\right)}$$$$\Rightarrow \frac{z^{″}cos\left(t\right)+sin\left(t\right)z^{\prime }}{{cos}^3\left(t\right)}-\frac{sin\left(t\right)}{{cos}^2\left(t\right)}.\frac{z^{\prime }\left(t\right)}{cos\left(t\right)}-\frac{z}{{cos}^2\left(t\right)}=sin\left(t\right)cos\left(t\right)$$$$\iff z^{″}-z=sin\left(t\right){cos}^3\left(t\right)=\frac{1}{2}sin\left(2t\right)\left(\frac{1+cos\left(2t\right)}{2}\right)$$$$\Rightarrow z^{″}-z=\frac{sin\left(2t\right)}{4}+\frac{sin\left(4t\right)}{8}$$$$z\left(t\right)={se}^t+{be}^{-t}forhomgeniusequation$$$$z_1=\frac{-sin\left(2t\right)}{20}-\frac{sin\left(4t\right)}{136}particularsolution$$$$z\left(t\right)={se}^t+{be}^{-t}-\frac{sin\left(2t\right)}{20}-\frac{sin\left(4t\right)}{136}$$$$t=arcsin\left(x\right)$$$$z\left(t\right)=z\left(arcsin\left(x\right)\right)=y\left(x\right)$$$$y\left(x\right)={se}^{arcsin\left(x\right)}+{be}^{-arcsin\left(x\right)}-\frac{x\sqrt{1-x^2}}{10}-\frac{x\sqrt{1-x^2}.(1-2x^2)}{34}$$

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