Question:
General solution of
\[
xp^2-yp+a=0,
\]
is \( \_\_\_\_ \), where
\[
p=\frac{dy}{dx}.
\]
General solution of
\[
xp^2-yp+a=0,
\]
is \( \_\_\_\_ \), where
\[
p=\frac{dy}{dx}.
\]
Show Hint
For Clairaut's equation \(y=px+f(p)\), the general solution is obtained by replacing \(p\) with constant \(c\).
Updated On: Jul 4, 2026
- \(y=cx+\dfrac{a}{c}\)
- \(y=cx-\dfrac{a}{c^2}\)
- \(y=cx+\dfrac{c^2}{a}\)
- \(y=x^2+\dfrac{a}{c^2}\)
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The Correct Option is A
Solution and Explanation
Concept:
The given equation is \[ xp^2-yp+a=0 \] where \[ p=\frac{dy}{dx} \] We rearrange it to express \(y\) in terms of \(x\) and \(p\).
Step 1: Rearrange the equation. \[ xp^2-yp+a=0 \] Move \(-yp\) to the other side: \[ yp=xp^2+a \] Divide by \(p\): \[ y=xp+\frac{a}{p} \]
Step 2: Recognize Clairaut's form. The equation \[ y=xp+f(p) \] is Clairaut's differential equation. Its general solution is obtained by putting \[ p=c \] where \(c\) is an arbitrary constant.
Step 3: Substitute \(p=c\). \[ y=xc+\frac{a}{c} \] or \[ y=cx+\frac{a}{c} \]
Step 4: Final answer. \[ \boxed{y=cx+\frac{a}{c}} \]
The given equation is \[ xp^2-yp+a=0 \] where \[ p=\frac{dy}{dx} \] We rearrange it to express \(y\) in terms of \(x\) and \(p\).
Step 1: Rearrange the equation. \[ xp^2-yp+a=0 \] Move \(-yp\) to the other side: \[ yp=xp^2+a \] Divide by \(p\): \[ y=xp+\frac{a}{p} \]
Step 2: Recognize Clairaut's form. The equation \[ y=xp+f(p) \] is Clairaut's differential equation. Its general solution is obtained by putting \[ p=c \] where \(c\) is an arbitrary constant.
Step 3: Substitute \(p=c\). \[ y=xc+\frac{a}{c} \] or \[ y=cx+\frac{a}{c} \]
Step 4: Final answer. \[ \boxed{y=cx+\frac{a}{c}} \]
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