Question:
The solution of the linear differential equation \( \dfrac{dy}{dx}+y=e^{-x} \), when \( x=0,\ y=1 \), is
The solution of the linear differential equation \( \dfrac{dy}{dx}+y=e^{-x} \), when \( x=0,\ y=1 \), is
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For a first-order linear differential equation \( \dfrac{dy}{dx}+P(x)y=Q(x) \), always find the integrating factor \( e^{\int P(x)\,dx} \). It turns the left side into a simple product derivative.
Updated On: Apr 28, 2026
- \( ye^{-x}=x-1 \)
- \( ye^{-x}=e^x-1 \)
- \( ye^x=x+1 \)
- \( ye^{-x}=e^x+1 \)
- \( ye^x=x-1 \)
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The Correct Option is C
Solution and Explanation
Step 1: Identify the equation as a linear differential equation.
The given equation is \[ \frac{dy}{dx}+y=e^{-x} \] This is in the standard linear form \[ \frac{dy}{dx}+P(x)y=Q(x) \] where \[ P(x)=1,\qquad Q(x)=e^{-x} \]
Step 2: Find the integrating factor.
The integrating factor is \[ I.F.=e^{\int P(x)\,dx}=e^{\int 1\,dx}=e^x \]
Step 3: Multiply the whole equation by the integrating factor.
Multiplying by \( e^x \), we get \[ e^x\frac{dy}{dx}+e^x y=e^x\cdot e^{-x} \] \[ e^x\frac{dy}{dx}+e^x y=1 \]
Step 4: Recognize the left side as a product derivative.
We know that \[ \frac{d}{dx}(ye^x)=e^x\frac{dy}{dx}+e^x y \] Therefore the equation becomes \[ \frac{d}{dx}(ye^x)=1 \]
Step 5: Integrate both sides.
Integrating with respect to \( x \), \[ \int \frac{d}{dx}(ye^x)\,dx=\int 1\,dx \] \[ ye^x=x+C \]
Step 6: Use the initial condition \( x=0,\ y=1 \).
Substitute \[ x=0,\qquad y=1 \] into \[ ye^x=x+C \] Then, \[ 1\cdot e^0=0+C \] \[ 1=C \] So the solution becomes \[ ye^x=x+1 \]
Step 7: Final conclusion.
Hence, the required solution is \[ \boxed{ye^x=x+1} \] Therefore, the correct option is \[ \boxed{(3)\ ye^x=x+1} \]
The given equation is \[ \frac{dy}{dx}+y=e^{-x} \] This is in the standard linear form \[ \frac{dy}{dx}+P(x)y=Q(x) \] where \[ P(x)=1,\qquad Q(x)=e^{-x} \]
Step 2: Find the integrating factor.
The integrating factor is \[ I.F.=e^{\int P(x)\,dx}=e^{\int 1\,dx}=e^x \]
Step 3: Multiply the whole equation by the integrating factor.
Multiplying by \( e^x \), we get \[ e^x\frac{dy}{dx}+e^x y=e^x\cdot e^{-x} \] \[ e^x\frac{dy}{dx}+e^x y=1 \]
Step 4: Recognize the left side as a product derivative.
We know that \[ \frac{d}{dx}(ye^x)=e^x\frac{dy}{dx}+e^x y \] Therefore the equation becomes \[ \frac{d}{dx}(ye^x)=1 \]
Step 5: Integrate both sides.
Integrating with respect to \( x \), \[ \int \frac{d}{dx}(ye^x)\,dx=\int 1\,dx \] \[ ye^x=x+C \]
Step 6: Use the initial condition \( x=0,\ y=1 \).
Substitute \[ x=0,\qquad y=1 \] into \[ ye^x=x+C \] Then, \[ 1\cdot e^0=0+C \] \[ 1=C \] So the solution becomes \[ ye^x=x+1 \]
Step 7: Final conclusion.
Hence, the required solution is \[ \boxed{ye^x=x+1} \] Therefore, the correct option is \[ \boxed{(3)\ ye^x=x+1} \]
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