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The order and degree the differential equation \( \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{2} y}{d x^{2}}} \)are pespectively. |
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Answer» Given differential equation is, \(\frac{d^2y}{dx^2}+\frac{dy}{dx}+x=\sqrt{1+\frac{d^2y}{dx^2}}\) By Squaring both sides, we get \((\frac{d^2y}{dx^2}+\frac{dy}{dx}+x)^2=1+\frac{d^2y}{dx^2}\) \(\Rightarrow\) \((\frac{d^2y}{dx^2})^ 2-\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2+2\,\frac{d^2y}{dx^2}\frac{dy}{dx}+2x\frac{d^2y}{dx^2}+2x\frac{dy}{dx}+x^2=1\) which is differential equation of order 2 and degree 2. |
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