Question:
If \(yf(x)=g(x)+c\) where \(c\) is an arbitrary constant such that \(\frac{dy}{dx}+yg(x)=f(x)\) and \([f(x)]^{2}-[g(x)]^{2}=1\), then \(\int f(x)\cdot g(x)dx=\) ...
If \(yf(x)=g(x)+c\) where \(c\) is an arbitrary constant such that \(\frac{dy}{dx}+yg(x)=f(x)\) and \([f(x)]^{2}-[g(x)]^{2}=1\), then \(\int f(x)\cdot g(x)dx=\) ...
Show Hint
Compare the two ways of obtaining y' to show f' = fg and g' = f^2.
Updated On: Nov 12, 2026
- \(g(x)+c\)
- \(-g(x)+c\)
- \(f(x)+c\)
- \(-f(x)+c\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
We have \(yf=g+c\) and \(y'+yg=f\). Differentiate the first relation: \(y'f+yf'=g'\).
Step 2: Eliminate y'
Multiply the differential equation by \(f\): \(y'f+ygf=f^{2}\).
Subtracting gives \(y(f'-gf)=g'-f^{2}\).
Step 3: Use the arbitrary constant
Since \(c\) is arbitrary, \(y\) varies freely, so both sides vanish: \(f'=fg\) and \(g'=f^{2}\).
This is consistent with \(f^{2}-g^{2}=1\), whose derivative gives \(2ff'-2gg'=0\).
Step 4: Integrate
\(\int f(x)g(x)dx=\int f'(x)dx=f(x)+c\).
Final Answer:
The integral equals \(f(x)+c\). \[ \boxed{\text{(C) }f(x)+c} \]
We have \(yf=g+c\) and \(y'+yg=f\). Differentiate the first relation: \(y'f+yf'=g'\).
Step 2: Eliminate y'
Multiply the differential equation by \(f\): \(y'f+ygf=f^{2}\).
Subtracting gives \(y(f'-gf)=g'-f^{2}\).
Step 3: Use the arbitrary constant
Since \(c\) is arbitrary, \(y\) varies freely, so both sides vanish: \(f'=fg\) and \(g'=f^{2}\).
This is consistent with \(f^{2}-g^{2}=1\), whose derivative gives \(2ff'-2gg'=0\).
Step 4: Integrate
\(\int f(x)g(x)dx=\int f'(x)dx=f(x)+c\).
Final Answer:
The integral equals \(f(x)+c\). \[ \boxed{\text{(C) }f(x)+c} \]
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