Question
Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}-{\text{y}}=\cos\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}-{\text{y}}=\cos\text{x}$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Q}$
Where P = -1 and $\text{Q}=\cos\text{x}$
$\therefore\text{ I}.\text{F}.=\text{e}^{\int{\text{P}\text{dx}}}$
$=\text{e}^{-\int\text{dx}}$
$=\text{e}^{-\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{-\text{x}},$ we get
$\text{e}^{-\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=\text{e}^{-\text{x}}\cos\text{x}$
$\Rightarrow\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{-\text{x}}\text{y}=\text{e}^{-\text{x}}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{-\text{x}}=\int \ \text{e}^{-\text{x}} \cos\text{x}\text{ dx} \ + \ \text{C}$
$\Rightarrow\text{ye}^{-\text{x}}=\text{I}+\text{C} \ ....(2)$
Here,
$\text{I}=\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}\ ..(3)$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin{\text{x}}-\int\big(-\text{e}^{-\text{x}}\sin\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}+\int {\text{e}^{-\text{x}}}\sin\text{x}\text{ dx}$
$\Rightarrow\text{I}= \text{e}^{-\text{x}}\sin \text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int[(-\text{e}^{-\text{x}})\times(-\cos\text{x})]\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x} - \text{e}^{-\text{x}}\cos\text{x} - \text{I}$ [From (3)]
$ \Rightarrow2\text{I}=\text{e}^{-\text{x}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x}-\cos{\text{x}})\ ...(4)$
From (2) and (4) we get
$\Rightarrow\text{y}\text{e}^{-\text{x}}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x} - \cos\text{x})+\text{C}$
$\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$
Hence, $\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$ is the requires solution.
$\frac{\text{dy}}{\text{dx}}-{\text{y}}=\cos\text{x}$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Q}$
Where P = -1 and $\text{Q}=\cos\text{x}$
$\therefore\text{ I}.\text{F}.=\text{e}^{\int{\text{P}\text{dx}}}$
$=\text{e}^{-\int\text{dx}}$
$=\text{e}^{-\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{-\text{x}},$ we get
$\text{e}^{-\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=\text{e}^{-\text{x}}\cos\text{x}$
$\Rightarrow\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{-\text{x}}\text{y}=\text{e}^{-\text{x}}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{-\text{x}}=\int \ \text{e}^{-\text{x}} \cos\text{x}\text{ dx} \ + \ \text{C}$
$\Rightarrow\text{ye}^{-\text{x}}=\text{I}+\text{C} \ ....(2)$
Here,
$\text{I}=\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}\ ..(3)$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin{\text{x}}-\int\big(-\text{e}^{-\text{x}}\sin\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}+\int {\text{e}^{-\text{x}}}\sin\text{x}\text{ dx}$
$\Rightarrow\text{I}= \text{e}^{-\text{x}}\sin \text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int[(-\text{e}^{-\text{x}})\times(-\cos\text{x})]\text{dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x}-\text{e}^{-\text{x}}\cos\text{x}-\int\text{e}^{-\text{x}}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\text{e}^{-\text{x}}\sin\text{x} - \text{e}^{-\text{x}}\cos\text{x} - \text{I}$ [From (3)]
$ \Rightarrow2\text{I}=\text{e}^{-\text{x}}(\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x}-\cos{\text{x}})\ ...(4)$
From (2) and (4) we get
$\Rightarrow\text{y}\text{e}^{-\text{x}}=\frac{\text{e}^{-\text{x}}}{2}(\sin\text{x} - \cos\text{x})+\text{C}$
$\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$
Hence, $\text{y}=\frac{1}{2}(\sin\text{x} - \cos\text{x}) +\text{C}\text{e}^{\text{x}}$ is the requires solution.
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