Question:
The solution of initial value problem: \(\frac{dy}{dx}=e^{3x+4y};\ y(0)=-\frac{1}{4}\) is
The solution of initial value problem: \(\frac{dy}{dx}=e^{3x+4y};\ y(0)=-\frac{1}{4}\) is
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Separate as \(e^{-4y}dy=e^{3x}dx\), integrate, then use \(y(0)=-\frac{1}{4}\) so that \(e^{-4y}=e\).
Updated On: Oct 1, 2026
- \(e^{-4y}+e^{3x}=3e\)
- \(3e^{-4y}+4e^{3x}=3e+4\)
- \(-4e^{-4y}+3e^{3x}=3e+4\)
- \(e^{-4y}-3e^{3x}=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We have a differential equation with a starting condition. We must find the particular solution that passes through \(x=0,\ y=-\frac{1}{4}\).
Step 2: Key Formula or Approach:
Write \(e^{3x+4y}=e^{3x}\cdot e^{4y}\). Then separate the variables and integrate. Use the condition to find the constant.
Step 3: Separate the variables.
\[ \frac{dy}{dx}=e^{3x}e^{4y} \]
\[ e^{-4y}dy=e^{3x}dx \]
Step 4: Integrate.
\[ -\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+C \]
Step 5: Use the initial condition.
Put \(x=0\) and \(y=-\frac{1}{4}\). Then \(e^{-4y}=e^{1}=e\) and \(e^{3x}=1\).
\[ -\frac{e}{4}=\frac{1}{3}+C \implies C=-\frac{e}{4}-\frac{1}{3} \]
Step 6: Simplify.
Substitute \(C\) back: \(-\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}-\frac{e}{4}-\frac{1}{3}\).
Multiply by 12: \(-3e^{-4y}=4e^{3x}-3e-4\).
Rearrange: \[ 3e^{-4y}+4e^{3x}=3e+4 \]
Step 7: Check the other options.
Test the point \((0,-\frac{1}{4})\) in each option. Option 1 gives \(e+1\), not \(3e\). Option 3 gives \(-4e+3\), not \(3e+4\). Option 4 gives \(e-3\), not 0. Only option 2 gives \(3e+4\).
Final Answer:
The solution is \(3e^{-4y}+4e^{3x}=3e+4\), option 2. \[ \boxed{3e^{-4y}+4e^{3x}=3e+4} \]
We have a differential equation with a starting condition. We must find the particular solution that passes through \(x=0,\ y=-\frac{1}{4}\).
Step 2: Key Formula or Approach:
Write \(e^{3x+4y}=e^{3x}\cdot e^{4y}\). Then separate the variables and integrate. Use the condition to find the constant.
Step 3: Separate the variables.
\[ \frac{dy}{dx}=e^{3x}e^{4y} \]
\[ e^{-4y}dy=e^{3x}dx \]
Step 4: Integrate.
\[ -\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+C \]
Step 5: Use the initial condition.
Put \(x=0\) and \(y=-\frac{1}{4}\). Then \(e^{-4y}=e^{1}=e\) and \(e^{3x}=1\).
\[ -\frac{e}{4}=\frac{1}{3}+C \implies C=-\frac{e}{4}-\frac{1}{3} \]
Step 6: Simplify.
Substitute \(C\) back: \(-\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}-\frac{e}{4}-\frac{1}{3}\).
Multiply by 12: \(-3e^{-4y}=4e^{3x}-3e-4\).
Rearrange: \[ 3e^{-4y}+4e^{3x}=3e+4 \]
Step 7: Check the other options.
Test the point \((0,-\frac{1}{4})\) in each option. Option 1 gives \(e+1\), not \(3e\). Option 3 gives \(-4e+3\), not \(3e+4\). Option 4 gives \(e-3\), not 0. Only option 2 gives \(3e+4\).
Final Answer:
The solution is \(3e^{-4y}+4e^{3x}=3e+4\), option 2. \[ \boxed{3e^{-4y}+4e^{3x}=3e+4} \]
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