The differential equation whose solution is A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $Ax^2 + By^2 = 1$. Since there are $2$ arbitrary constants ($A$ and $B$),we differentiate the equation twice. First differentia

Vedclass · Accepted Answer

degree $1$ and order $2$

Vedclass · Answer

(A) Given the equation $Ax^2 + By^2 = 1$. Since there are $2$ arbitrary constants ($A$ and $B$),we differentiate the equation twice. First differentiation with respect to $x$: $2Ax + 2Byy' = 0$,which simplifies to $Ax + Byy' = 0$. Second differentiation with respect to $x$: $A + B(y')^2 + Byy'' = 0$. From the first derivative,$A = -Byy'/x$. Substituting $A$ into the second derivative: $-Byy'/x + B(y')^2 + Byy'' = 0$. Dividing by $B$ (assuming $B 
eq 0$): $-yy'/x + (y')^2 + yy'' = 0$. Multiplying by $x$: $-yy' + x(y')^2 + xyy'' = 0$. The highest order derivative present is $y''$,so the order is $2$. The power of the highest order derivative $y''$ is $1$,so the degree is $1$.

The differential equation whose solution is $Ax^2 + By^2 = 1$,where $A$ and $B$ are arbitrary constants,is of:

Similar Questions

The degree of the differential equation $\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)$ is

The order of the differential equation whose solution is $y = a\cos x + b\sin x + c{e^{ - x}}$ is

The differential equation $x(\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y = x^2$ is of

The order and degree of the differential equation $3 - (\frac{d^3y}{dx^3})^{2/3} = (\frac{dy}{dx})^5$ are respectively

The order and degree of the differential equation $1+(\frac{dy}{dx})^2=\sqrt{\frac{d^2y}{dx^2}}$ are . . . . . . and . . . . . . respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module