Question
Solve the following differential equation:$4\frac{\text{dy}}{\text{dx}}+8\text{y}=5\text{e}^{-3\text{x}}$
✓
Answer
$4\frac{\text{dy}}{\text{dx}}+8\text{y}=5\text{e}^{-3\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+2\text{y}=\frac{5}4\text{e}^{-3\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=2$
$\text{Q}=\frac{5}4\text{e}^{-3\text{x}}$
$\therefore$ 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
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\frac{5}4\text{e}^{2\text{x}}\text{e}^{-3\text{x}}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\frac{5}4\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\frac{5}4\int\text{e}^{-\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=-\frac{5}4\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$ is the required solution.
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+2\text{y}=\frac{5}4\text{e}^{-3\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=2$
$\text{Q}=\frac{5}4\text{e}^{-3\text{x}}$
$\therefore$ 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
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\frac{5}4\text{e}^{2\text{x}}\text{e}^{-3\text{x}}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\frac{5}4\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\frac{5}4\int\text{e}^{-\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=-\frac{5}4\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$ is the required solution.
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