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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}$
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Answer

We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=4\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=4\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}4\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\int\text{x}\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\int\text{e}^{2\text{x}}\text{dx}-4\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{2\text{x}}\text{dx}\Big]\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=4\text{x}\frac{\text{e}^{2\text{x}}}2-4\times\frac{1}2\int\text{e}^{2\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-4\times\frac{1}4\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=2\text{x}\ \text{e}^{2\text{x}}-\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=(2\text{x}-1)\text{e}^{2\text{x}}+\text{C}\text{e}^{-2\text{x}}$ is the required solution.

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