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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0,\text{y}(0)=1,\text{y}(0)=3$
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 - ex + 2e2x is the solution of the given equation.

put x = 0 in equation (i),

y = e0 + e0

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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