Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}\cos\text{x}$
$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}\cos\text{x}$
✓
Answer
Here, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=\cos\text{x}$ It is a linear differential equation. Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{2}{\text{x}},\text{Q}=\cos\text{x}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{2\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{2\log|\text{x}|}$ $=\text{x}^2$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.}=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}(\text{x}^2)=\int\cos\text{x}(\text{x}^2)\text{dx + C}$ $\text{yx}^2=\int\text{x}^2\cos\text{xdx + C}$ $=\text{x}^2\int\cos\text{x}-\int(2\text{x}\times\int\cos\text{xdx})\text{dx + C}$ Usind integration by parts $\text{yx}^2=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx + C}$ $=\text{x}^2\sin\text{x}-2\big[\text{x}\times\int\sin\text{xdx}-\int(1\times\int\sin\text{xdx})\text{dx}\big]+\text{C}$ $=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int\cos\text{xdx + C}$ $\text{yx}^2=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x + C}$ $\text{y}=\sin\text{x}+\frac{2}{\text{x}}\cos\text{x}-\frac{2}{\text{x}^2}\sin\text{x}+\frac{\text{C}}{\text{x}^2}$Need a full question paper?
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