The differential equation whose solution is A | vedclass

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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

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degree $1$ and order $2$

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(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:

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