Question:
If the general solution of the differential equation
\[
\cos^2x\frac{dy}{dx}+y=\tan x
\]
is
\[
y=\tan x-1+Ce^{-\tan x}
\]
and it satisfies
\[
y\left(\frac{\pi}{4}\right)=1,
\]
then \(C=\)
If the general solution of the differential equation
\[
\cos^2x\frac{dy}{dx}+y=\tan x
\]
is
\[
y=\tan x-1+Ce^{-\tan x}
\]
and it satisfies
\[
y\left(\frac{\pi}{4}\right)=1,
\]
then \(C=\)
Show Hint
To find the arbitrary constant in a differential equation, substitute the given initial condition directly into the general solution.
Updated On: Jun 22, 2026
- \(e\)
- \(1\)
- \(-1\)
- \(\frac{1}{e}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the given general solution.
Given, \[ y=\tan x-1+Ce^{-\tan x} \] Also, \[ y\left(\frac{\pi}{4}\right)=1 \] Substitute \[ x=\frac{\pi}{4} \] in the solution.
Step 2: Evaluate trigonometric values.
Since, \[ \tan\frac{\pi}{4}=1 \] we get \[ 1=1-1+Ce^{-1} \] \[ 1=\frac{C}{e} \]
Step 3: Solve for \(C\).
Multiplying both sides by \(e\), \[ C=e \]
Step 4: Final conclusion.
Therefore, \[ \boxed{e} \]
Given, \[ y=\tan x-1+Ce^{-\tan x} \] Also, \[ y\left(\frac{\pi}{4}\right)=1 \] Substitute \[ x=\frac{\pi}{4} \] in the solution.
Step 2: Evaluate trigonometric values.
Since, \[ \tan\frac{\pi}{4}=1 \] we get \[ 1=1-1+Ce^{-1} \] \[ 1=\frac{C}{e} \]
Step 3: Solve for \(C\).
Multiplying both sides by \(e\), \[ C=e \]
Step 4: Final conclusion.
Therefore, \[ \boxed{e} \]
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