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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
verify that $\text{y}=\text{e}^{\text{m}\cos^{-1}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{m}^2\text{y}=0$
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Answer

We have,
$\text{y}=\text{e}^{\text{m}\cos^{-1}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{me}^{\text{m}^{\cos^{-1}}}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{me}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^3}}\ ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{me}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-\text{m})\Bigg[\frac{\sqrt{1-\text{x}^2}\text{me}^{\text{m}^{\cos^{-1}}}\Big(-\frac{1}{\sqrt{1-\text{x}}}\Big)-\text{e}^{\text{m}^{\cos^{-1}}}\text{x}\frac{1}{2}\Big(-\frac{2\text{x}}{\sqrt{1-\text{x}^2}}\Big)}{(1-\text{x}^2)}\Bigg]$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-\text{m})\Big[-\text{me}^{\text{m}^{\cos^{-1}}}\text{x}+\frac{\text{xe}^{\text{m}^{\cos^{-1}}}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{m}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{m}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{m}^2\text{y}=0$
Hence, the given function is the solution to the given differential equation.

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