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Differential Equations — Maths STD 12 Science — Question

Maharashtra 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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