Answer :

ax^{2} + by^{2} = 1 will be solution of differential equation x (yy_{2} + y_{1}^{2}) = yy_{1} if after solving,

that is integrating the differential equation we again get ax^{2} + by^{2} = 1.

We will use and

The given differential equation is of second order.

We can also say that if after differentiating ax^{2} + by^{2} = 1 twice we will get the given differential equation then it is solution of given differential equation

Differentiate ax^{2} + by^{2} = 1 with respect to x

⇒ 2ax + 2byy_{1} = 0

⇒ ax + byy_{1} = 0 …(i)

Differentiating again with respect to x

Using the product rule (UV)^{’}= UV^{’} + VU^{’} we get,

a + b (y_{1}y_{1} + yy_{2}) = 0

⇒ a + b (y_{1}^{2} + yy_{2}) = 0

⇒ a = -b (y_{1}^{2} + yy_{2})

We have to eliminate the constants a and b hence substitute value of a in (i)

⇒ -b (y_{1}^{2} + yy_{2}) x + byy_{1} = 0

Take out b common,

⇒ - xy_{1}^{2} – xyy_{2} + yy_{1} = 0

⇒ yy_{1} = xy_{1}^{2} + xyy_{2}

⇒ yy_{1} = x (y_{1}^{2} + yy_{2})

Hence after differentiating we get the required differential equation.

Hence ax^{2} + by^{2} = 1 is a solution of the differential equation x (yy_{2} + y_{1}^{2}) = yy_{1}.

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