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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