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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}$
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Answer

The given differential equation is $\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}.$ This is in the form of $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}$ (where p = 2 and Q = sin x). $\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int2\text{dx}}=\text{e}^{2\text{x}}.$The solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F})\text{dx}+\text{C}$ $\Rightarrow\ \text{ye}^{2\text{x}}=\int\sin\text{x}\cdot\text{e}^{2\text{x}}\text{dx}+\text{C}\ \ ....{(1)}$ $\text{Let}\ I=\int\sin\text{x}\cdot\text{e}^{2\text{x}}.$ $\Rightarrow\ I=\sin\text{x}\cdot\int\text{e}^{2\text{x}}\text{dx}-\int\bigg(\frac{\text{d}}{\text{dx}}(\sin\text{x})\cdot\int\text{e}^{2\text{x}}\text{dx}\bigg)\text{dx}$ $\Rightarrow\ I=\sin\text{x}\cdot\frac{\text{e}^{2\text{x}}}{2}-\int\bigg(\cos\text{x}\cdot\frac{\text{e}^{2\text{x}}}{2}\bigg)\text{dx}$ $\Rightarrow\ I=\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}-\frac{1}{2}\bigg[\cos\text{x}\cdot\int\text{e}^{2\text{x}}-\int\bigg(\frac{\text{d}}{\text{dx}}(\cos\text{x})\cdot\int\text{e}^{2\text{x}}\text{dx}\bigg)\text{dx}\bigg]$ $\Rightarrow\ I=\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}-\frac{1}{2}\bigg[\cos\text{x}\cdot\frac{\text{e}^{2\text{x}}}{2}-\int\bigg[(-\sin\text{x})\cdot\frac{\text{e}^{2\text{x}}}{2}\bigg]\text{dx}\bigg]$ $\Rightarrow\ I=\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}-\frac{\text{e}^{2\text{x}}\cos\text{x}}{4}-\frac{1}{4}\int\big(\sin\text{x}.\text{e}^{2\text{x}}\big)\text{dx}$ $\Rightarrow\ I=\frac{\text{e}^{2\text{x}}}{4}(2\sin\text{x}-\cos\text{x})-\frac{1}{4}I$ $\Rightarrow\ \frac{5}{4}I=\frac{\text{e}^{2\text{x}}}{4}(2\sin\text{x}-\cos\text{x})$ $\Rightarrow\ I=\frac{\text{e}^{2\text{x}}}{5}(2\sin\text{x}-\cos\text{x})$ Therefore, equation (1) becomes: $\text{y}^{2\text{x}}=\frac{\text{e}^{2\text{x}}}{5}(2\sin\text{x}-\cos\text{x})+\text{C}$ $\Rightarrow\ \text{y}=\frac{1}{5}(2\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-2\text{x}}$ This is the required general solution of the given differential equation.

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