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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 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 = ex + e-x satisfies the given differential equation.

Also, when x = 0; = e0 + e0 = 1 + 1, i.e. y(0) = 2.

And, when x = 0; y1 = e0 - e0 = 1 - 1, i.e. y'(0) = 0

Hence, y = ex + e-x is the solution to the given initial value problem.

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