Question:
The general solution of $xp^{2} - 2yp + ax = 0$ is where $p = \frac{dy}{dx}$.
The general solution of $xp^{2} - 2yp + ax = 0$ is where $p = \frac{dy}{dx}$.
Show Hint
If an equation is easily solvable for $y$, differentiate with respect to $x$ to create a new equation in $p$ and $x$.
Updated On: Jul 3, 2026
- $2x = cy + ac^{2}$
- $2y = cx^{2} + \frac{a}{c}$
- $y = cx^{2} + ac^{2}$
- $y = cx^{2} + \frac{a}{c}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Concept
This is a differential equation of the first order but not of the first degree, solvable for $y$.
Step 2: Meaning
Expressing $y$ in terms of $x$ and $p$: $2yp = xp^{2} + ax \Rightarrow 2y = xp + \frac{ax}{p}$.
Step 3: Analysis
Differentiating with respect to $x$ and replacing $dy/dx$ with $p$ allows us to find a relation between $p$ and $x$. Integrating that relation gives $p$ as a function of $x$ and a constant $c$.
Step 4: Conclusion
Substituting the expression for $p$ back into the equation for $y$ simplifies to the general solution: $2y = cx^{2} + \frac{a}{c}$. Final Answer: (B)
This is a differential equation of the first order but not of the first degree, solvable for $y$.
Step 2: Meaning
Expressing $y$ in terms of $x$ and $p$: $2yp = xp^{2} + ax \Rightarrow 2y = xp + \frac{ax}{p}$.
Step 3: Analysis
Differentiating with respect to $x$ and replacing $dy/dx$ with $p$ allows us to find a relation between $p$ and $x$. Integrating that relation gives $p$ as a function of $x$ and a constant $c$.
Step 4: Conclusion
Substituting the expression for $p$ back into the equation for $y$ simplifies to the general solution: $2y = cx^{2} + \frac{a}{c}$. Final Answer: (B)
Was this answer helpful?
0
0
Top AP ECET BSc Mathematics Mathematics Questions
- The integrating factor of $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is}
- If $y = y(x)$ is the solution of $\left(\frac{2 + \sin x}{y + 1}\right) \frac{dy}{dx} + \cos x = 0$ with $y(0) = 1$, then $y\left(\frac{\pi}{2}\right) = $
- The solution of $y^{4}dx + 2xy^{3}dy = \frac{ydx - xdy}{x^{3}y^{3}}$ is}
- Which of the following is an integrating factor of $(x^{2} + y^{2} + 2x)dx + 2ydy = 0$?
- The differential equation $\frac{dy}{dx} = \frac{-(x + x^{8} + py^{2})}{y^{8} - y + qxy}$ is exact if}
View More Questions
Top AP ECET BSc Mathematics Differential equations Questions
- The integrating factor of $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is}
- If $y = y(x)$ is the solution of $\left(\frac{2 + \sin x}{y + 1}\right) \frac{dy}{dx} + \cos x = 0$ with $y(0) = 1$, then $y\left(\frac{\pi}{2}\right) = $
- The solution of $y^{4}dx + 2xy^{3}dy = \frac{ydx - xdy}{x^{3}y^{3}}$ is}
- Which of the following is an integrating factor of $(x^{2} + y^{2} + 2x)dx + 2ydy = 0$?
- The differential equation $\frac{dy}{dx} = \frac{-(x + x^{8} + py^{2})}{y^{8} - y + qxy}$ is exact if}
View More Questions
Top AP ECET BSc Mathematics Questions
- The integrating factor of $x \cos x \frac{dy}{dx} + (x \sin x + \cos x) y = 1$ is}
- If $y = y(x)$ is the solution of $\left(\frac{2 + \sin x}{y + 1}\right) \frac{dy}{dx} + \cos x = 0$ with $y(0) = 1$, then $y\left(\frac{\pi}{2}\right) = $
- The solution of $y^{4}dx + 2xy^{3}dy = \frac{ydx - xdy}{x^{3}y^{3}}$ is}
- Which of the following is an integrating factor of $(x^{2} + y^{2} + 2x)dx + 2ydy = 0$?
- The differential equation $\frac{dy}{dx} = \frac{-(x + x^{8} + py^{2})}{y^{8} - y + qxy}$ is exact if}
View More Questions