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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0,\text{y}(0)=1,\text{y}(0)=3$
Function $\text{y}=\text{e}^\text{x}+\text{e}^{2\text{x}}$
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Answer

$\text{y}=\text{e}^{\text{x}}+\text{e}^{2\text{x}} ...(\text{i})$ Differentiating it with respect to $x, \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}+2\text{e}^{2\text{x}} ...\text{(ii)}$
Again, differentiating it with respect to $x, \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{e}^{\text{x}}+4\text{e}^{2\text{x}}$
$=(3-2)\text{e}^{\text{x}}+(6-2)\text{e}^{2\text{x}}$
$=3\text{e}^{\text{x}}+6\text{e}^{2\text{x}}-2\text{e}^{\text{x}}-2\text{e}^{2\text{x}}$
$=3(\text{e}^\text{x}+2\text{e}^{2\text{x}})-2 (\text{e}^{\text{x}}+\text{e}^{2\text{x}})$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=3\frac{\text{dy}}{\text{dx}}-2\text{y}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
It is the given equation,
so $y - e^x + 2e^{2x}$ is the solution of the given equation. put $x = 0$ in equation $(i),y = e^{0 }+ e^0$
$y = 1 + 1$
$y = 2$
so,
$y(0) = 2$
put $x - 0$ in equation $(ii),$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{0}+2\text{e}^{0} y' = 1 + 2$
$y' = 3$
so, $y'(0) = 3$

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