The solution of $\frac{d y}{d x}+y=e^x$ is

The solution of $\frac{d y}{d x}+y=e^x$ is
  1. $2 y=e^{2 x}+C$
  2. $2 y e^x=e^x+C$
  3. $2 y e^x=e^{2 x}+C$
  4. $2 y e^{2 x}=2 e^x+C$

Solution

We have, $\frac{d y}{d x}+y=e^x$ On comparing with $\frac{d y}{d x}+P y=Q$ Here, $P=1$ and $Q=e^x$ Now, IF $=e^{\int 1 d x}=e^x$ Complete solution is $\begin{array}{rlrl} & y \cdot e^x & =\int e^x \cdot e^x d x+C^{\prime} \\ \Rightarrow \quad & y e^x & =\frac{e^{2 x}}{2}+C^{\prime} \\ \Rightarrow & & y y e^x & =e^{2 x}+C\end{array}$ $\left[\because 2 C^{\prime}=C\right]$

Asked in: AP EAMCET 2001

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