Question:
Find the particular solution of the differential equation \( (1 + e^{2x})dy + (1 + y^2)e^x dx = 0 \), given that \( y(1) = 0 \).
Find the particular solution of the differential equation \( (1 + e^{2x})dy + (1 + y^2)e^x dx = 0 \), given that \( y(1) = 0 \).
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Always check if simple variable separation is possible before attempting more complex methods.
Ensure substitution is correctly substituted back before applying initial conditions.
Ensure substitution is correctly substituted back before applying initial conditions.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Variable Separable Method: Grouping \( y \) terms with \( dy \) and \( x \) terms with \( dx \).
• Particular Solution involves finding the integration constant \( C \) using boundary conditions.
Step 1: Separate the variables
\[ (1 + e^{2x})dy = -(1 + y^2)e^x dx \] \[ \frac{dy}{1 + y^2} = -\frac{e^x}{1 + (e^x)^2} dx \]
Step 2: Integrate both sides
\[ \int \frac{dy}{1 + y^2} = -\int \frac{e^x}{1 + (e^x)^2} dx \] For the RHS, put \( e^x = t \), so \( e^x dx = dt \). \[ \tan^{-1} y = -\int \frac{dt}{1 + t^2} = -\tan^{-1} t + C \] \[ \tan^{-1} y = -\tan^{-1} e^x + C \implies \tan^{-1} y + \tan^{-1} e^x = C \]
Step 3: Use boundary condition to find \( C \)
Given \( y(1) = 0 \), substitute \( x = 1, y = 0 \): \[ \tan^{-1} 0 + \tan^{-1} e^1 = C \] \[ 0 + \tan^{-1} e = C \implies C = \tan^{-1} e \]
Step 4: Write the particular solution
\[ \tan^{-1} y + \tan^{-1} e^x = \tan^{-1} e \]
• Variable Separable Method: Grouping \( y \) terms with \( dy \) and \( x \) terms with \( dx \).
• Particular Solution involves finding the integration constant \( C \) using boundary conditions.
Step 1: Separate the variables
\[ (1 + e^{2x})dy = -(1 + y^2)e^x dx \] \[ \frac{dy}{1 + y^2} = -\frac{e^x}{1 + (e^x)^2} dx \]
Step 2: Integrate both sides
\[ \int \frac{dy}{1 + y^2} = -\int \frac{e^x}{1 + (e^x)^2} dx \] For the RHS, put \( e^x = t \), so \( e^x dx = dt \). \[ \tan^{-1} y = -\int \frac{dt}{1 + t^2} = -\tan^{-1} t + C \] \[ \tan^{-1} y = -\tan^{-1} e^x + C \implies \tan^{-1} y + \tan^{-1} e^x = C \]
Step 3: Use boundary condition to find \( C \)
Given \( y(1) = 0 \), substitute \( x = 1, y = 0 \): \[ \tan^{-1} 0 + \tan^{-1} e^1 = C \] \[ 0 + \tan^{-1} e = C \implies C = \tan^{-1} e \]
Step 4: Write the particular solution
\[ \tan^{-1} y + \tan^{-1} e^x = \tan^{-1} e \]
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