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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0,\text{y}(0)=2,\text{y}'(0)=0$
Function $\text{y}=\text{e}^\text{x}+\text{e}^{-\text{x}}$
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Answer

We have $\text{y}=\text{e}^{\text{x}}+\text{e}^{\text{-x}} ...(1)$
Differentiating both sides of $(1)$ with respect to $x,$
we get $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}-\text{e}^{\text{x}} ...(2)$
Differentiating both sides of $(2)$ with respect to $x,$
we get $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{e}^{\text{x}}+\text{e}^{\text{-x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y} [$Using $(1)]$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
It is the given differential equation.
Therefore, $y = e^{x }+ e^{-x}$ satisfies the given differential equation.
Also, when $x = 0; = e^{0 }+ e^{0 }= 1 + 1,$ i.e. $y(0) = 2.$
And, when $x = 0; y_{1 }= e^{0 }- e^{0 }= 1 - 1,$ i.e. $y'(0) = 0$
Hence, $y = e^{x }+ e^{-x }$ is the solution to the given initial value problem.

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