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Differential Equations — Maths STD 12 Science — Question

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

We have,
$\text{y}=\text{ce}^{\tan^{-1}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{ce}^{\tan^{1}\text{x}}\frac{1}{1+\text{x}^2}\ ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{(1+\text{x}^2)\text{e}^{\tan^{1}}\text{x}\frac{1}{1+\text{x}^2}-\text{e}^{\tan^{1}}\text{x}(2\text{x})}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{\text{e}^{\tan^{-1}}\text{x}-2\text{xe}^{\tan^{-1}}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}\frac{(1-2\text{x})\text{e}^{\tan^{-1}}\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{c}(1-2\text{x})\frac{\text{e}^{\tan^{-1}}}{(1+\text{x}^2)}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=(1-2\text{x})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}-1)\frac{\text{dy}}{\text{dx}}=0$
Hence, the given function is the solution to the given differential equation.

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