Verify that ax² + by² = 1 is a solution of the differential equation x (yy₂ + y₁²) = yy₁

ax² + by² = 1 will be solution of differential equation x (yy₂ + y₁²) = yy₁ if after solving,

that is integrating the differential equation we again get ax² + by² = 1.

We will use and

The given differential equation is of second order.

We can also say that if after differentiating ax² + by² = 1 twice we will get the given differential equation then it is solution of given differential equation

Differentiate ax² + by² = 1 with respect to x

⇒ 2ax + 2byy₁ = 0

⇒ ax + byy₁ = 0 …(i)

Differentiating again with respect to x

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

a + b (y₁y₁ + yy₂) = 0

⇒ a + b (y₁² + yy₂) = 0

⇒ a = -b (y₁² + yy₂)

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

⇒ -b (y₁² + yy₂) x + byy₁ = 0

Take out b common,

⇒ - xy₁² – xyy₂ + yy₁ = 0

⇒ yy₁ = xy₁² + xyy₂

⇒ yy₁ = x (y₁² + yy₂)

Hence after differentiating we get the required differential equation.

Hence ax² + by² = 1 is a solution of the differential equation x (yy₂ + y₁²) = yy₁.