Question

Solve $16x^2+2p^{2}y-p^{3}x=0$ and mark the correct answer.

• A. $2+C^2{y}^2-C^3{x}^2=0$
• B. $4+C^{2}y-C^3{x}^2=5$
• C. $2+C^{3}y-C^2{x}^2=0$
• D. $2+C^{2}y-C^3{x}^2=0$

Answer 1

The correct option is D $2+C^{2}y-C^3{x}^2=0$

Given equation is $16x^2+2p^{2}y-p^{3}x=0$

We can rewrite the equation as
$2y=px-16\frac{x^2}{p^2}$
Differentiate with respect to $x$, we get
$2p=p+x\frac{dp}{dx}-\frac{32x}{p^2}+\frac{32x^2}{p^3}\frac{dp}{dx}$
$p(p^3+32x)-x(p^3+32x)\frac{dp}{dx}=0$
$(p^3+32x)(p-x\frac{dp}{dx})=0$
This equation is satified when $p^3+32x=0$ or $p-x\frac{dp}{dx}=0$
Then we obtain only
$\frac{dp}{p}=\frac{dx}{x}$
Integrating both the sides, we get
$\ln \left(p\right)=\ln \left(x\right)+\ln \left(C\right)$
$\ln \left(p\right)=\ln \left(xC\right)$
Take exponential on both the sides, we get
$p=xC$
Substituting the value in an given differential equation, we get
$16x^2+2C^2{x}^{2}y-C^3{x}^4=0$
Replacing $C$ by $2c$, we get
$2+c^{2}y-c^3{x}^2=0$ which is a primitive.