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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=1,\text{y}(0)=2$
Function $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$
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Answer

We have,
$\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}\ ...(1)$
Differentiating both sides of $(1)$ with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+\text{e}^{\text{x}}+\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+2\text{e}^{\text{x}}...(2)$
Differentiating both sides of $(2)$ with respect to $x,$ we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{xe}^{\text{x}}+\text{e}^{\text{x}}+2\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{xe}^{\text{x}}+3\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2(\text{xe}^{\text{x}}+2\text{e}^{\text{x}})(\text{xe}^{\text{x}}+\text{e}^{\text{x}})$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2\frac{\text{dy}}{\text{dx}}-\text{y} [$Using$(1)$ and $(2)]$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
It is the given differential equation.
Thus, $y = xe^x + e^x$
satisfies the given differential equation.
Also, when $x = 0, y = 0 + 1 = 1,$ i.e. $y(0) = 1$
And, when $x = 0, y\ ' = 0 + 2 = 2,$ i.e. $y\ '(0) = 2$
Hence, $y = xe^x + e^x$ is the solution to the given initialvalue problem.

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