Show that Ax^2+By^2=1 is a solution of the differential equation x y=x d^2y dx^2 + ( dx dy )^2 =y dy dx .

We have, Ax^2+By^2=1 ...(1) Differentiating it with respect in x 2Ax+2By dy dx =0 y dy dx = -2Ax 2B y dy dx = Ax B Differentiating it with respect in x y d^2y dx^2 + ( dy dx )^2=- A B y d^2y dx^2 + ( dy dx )^2= y x dy dx Using equation (1) x y d^2y dx^2 + ( dy dx )^2 =y dy dx

Show that Ax^2+By^2=1 is a solution of the differential equation …

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Question

Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$

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Answer

We have,

$\text{Ax}^2+\text{By}^2=1\ ...(1)$

Differentiating it with respect in x

$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$

$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$

$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$

Differentiating it with respect in x

$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$

$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$

Using equation (1)

$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$

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