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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations2 Marks
Question
Verify that the function $y=e^{x}(a \cos x+b \sin x)$ (implicit or explicit) is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$
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Answer

It is given that $y = e^x(a. \cos x + b \sin x) = a.e^x \cos x + b.e^x \sin x$
Now, differentiating both sides w.r.t. x, we get,
$\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x} \cos x\right)+b \frac{d}{d x}\left(e^{x} \sin x\right)$
$\Rightarrow \frac{d y}{d x}=a\left(e^{x} \cos x-e^{x} \sin x\right)+b \cdot\left(e^{x} \sin x+e^{x} \cos x\right)$
$\Rightarrow \frac{d y}{d x}=(a+b) e^{x} \cos x+(b-a) e^{x} \sin x$
Now, again differentiating both sides w.r.t. x, we get,
$\frac{d^{2} y}{d x^{2}}=(a+b) \cdot \frac{d}{d x}\left(e^{x} \cos x\right)+(b-a) \frac{d}{d x}\left(e^{x} \sin x\right)$
= $(a+b) \cdot\left[e^{x} \cos x-e^{x} \sin x\right]+(b-a)\left[e^{x} \sin x+e^{x} \cos x\right]$
= $\mathrm{e}^{\mathrm{x}}[\mathrm{a.cosx}-\mathrm{a.sinx}+\mathrm{b.cosx}-\mathrm{b.sinx}+\mathrm{b.sinx}+\mathrm{b.cosx}-\mathrm{a.sinx}-\mathrm{a.cosx}]$
= $\left[2 \mathrm{e}^{\mathrm{x}}(\mathrm{b.cosx}-\mathrm{a.sinx})\right]$
Now, Substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the given differential equations, we get,
LHS = $=\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y$
$= 2e^x(b \cos x - a \sin x) -2e^x[(a + b) \cos x + (b - a ) \sin x] + 2e^x(a \cos x + b \sin x)$
$=e^x[(2b \cos x - 2a \sin x) - (2a \cos x + 2b \cos x) - (2b \sin x - 2a \sin x) + (2a \cos x + 2b \sin x)]$
$= e^x[(2b - 2a - 2b + 2a) \cos x] + e^x[(-2a - 2b + 2a + 2b) \sin x]$
$= 0 =$ RHS.
Therefore, the given function is the solution of the corresponding differential equation.

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