Q. 105.0( 1 Vote )

# Verify that ax<su

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.

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