Question
Fill in the blanks.
General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ is _________.
General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ is _________.
✓
Answer
General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ is $\text{y}=\frac{1}{2}(\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-\text{x}}.$Solution:
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ Which is of the from $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{I.F.}=\int\text{e}^{1\text{dx}}=\text{e}^{\text{x}}$ So, the general solution is $\text{y}.\text{e}^{\text{x}}=\int\text{e}^{\text{x}}\sin\text{x dx}+\text{C}$ $\Rightarrow\text{y}.\text{e}^{\text{x}}=\frac{1}{2}\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})+\text{C}$ $\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-\text{x}}$
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\sin\text{x}$ Which is of the from $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{I.F.}=\int\text{e}^{1\text{dx}}=\text{e}^{\text{x}}$ So, the general solution is $\text{y}.\text{e}^{\text{x}}=\int\text{e}^{\text{x}}\sin\text{x dx}+\text{C}$ $\Rightarrow\text{y}.\text{e}^{\text{x}}=\frac{1}{2}\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})+\text{C}$ $\Rightarrow\text{y}=\frac{1}{2}(\sin\text{x}-\cos\text{x})+\text{C}\text{e}^{-\text{x}}$
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