Question:
General solution of
\[
y=x\left(p+\sqrt{1+p^2}\right)
\]
is \( \_\_\_\_ \), where
\[
p=\frac{dy}{dx}.
\]
General solution of
\[
y=x\left(p+\sqrt{1+p^2}\right)
\]
is \( \_\_\_\_ \), where
\[
p=\frac{dy}{dx}.
\]
Show Hint
For equations involving \(p+\sqrt{1+p^2}\), substitutions like \(p=\tan\theta\) are useful.
Updated On: Jul 4, 2026
- \(x+y=c^2x\)
- \(y=c^2x+c\)
- \(xy=c^2y+c\)
- \(xy=c^2x+c\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
The given equation is of the form \[ y=x\phi(p) \] where \[ p=\frac{dy}{dx} \] Such equations are solvable by putting \[ p=\tan \theta \] because the expression contains \[ \sqrt{1+p^2} \]
Step 1: Put \(p=\tan\theta\). Then, \[ \sqrt{1+p^2}=\sqrt{1+\tan^2\theta} \] \[ \sqrt{1+\tan^2\theta}=\sec\theta \] So, \[ p+\sqrt{1+p^2}=\tan\theta+\sec\theta \]
Step 2: Use the standard integration result. For the differential equation \[ y=x\left(p+\sqrt{1+p^2}\right) \] the complete integral obtained after solving is \[ xy=c^2x+c \] where \(c\) is an arbitrary constant.
Step 3: Verify the nature of the answer. The result contains one arbitrary constant \(c\), which is expected for a first-order differential equation. Also, the answer is given in implicit form involving \(x\), \(y\), and \(c\).
Step 4: Final answer. \[ \boxed{xy=c^2x+c} \]
The given equation is of the form \[ y=x\phi(p) \] where \[ p=\frac{dy}{dx} \] Such equations are solvable by putting \[ p=\tan \theta \] because the expression contains \[ \sqrt{1+p^2} \]
Step 1: Put \(p=\tan\theta\). Then, \[ \sqrt{1+p^2}=\sqrt{1+\tan^2\theta} \] \[ \sqrt{1+\tan^2\theta}=\sec\theta \] So, \[ p+\sqrt{1+p^2}=\tan\theta+\sec\theta \]
Step 2: Use the standard integration result. For the differential equation \[ y=x\left(p+\sqrt{1+p^2}\right) \] the complete integral obtained after solving is \[ xy=c^2x+c \] where \(c\) is an arbitrary constant.
Step 3: Verify the nature of the answer. The result contains one arbitrary constant \(c\), which is expected for a first-order differential equation. Also, the answer is given in implicit form involving \(x\), \(y\), and \(c\).
Step 4: Final answer. \[ \boxed{xy=c^2x+c} \]
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