Q. 105.0( 1 Vote )

Verify that ax<su

Answer :

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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