Q. 2 B3.6( 8 Votes )

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

y = ex (a cos x + b sin x) :

Answer :

It is given that y = ex(acosx + bsinx) = aexcosx + bexsinx


Now, differentiating both sides w.r.t. x, we get,





Now, again differentiating both sides w.r.t. x, we get,






Now, Substituting the values of ’ and in the given differential equations, we get,


LHS =


=2ex(bcosx – asinx) -2ex[(a + b)cosx + (b –a ) sinx] + 2ex(acosx + bsinx)


=ex[(2bcosx – 2asinx) - (2acosx + 2bcosx) - (2bsinx – 2asinx) + (2acosx + 2bsinx)]


= ex[(2b – 2a – 2b + 2a)cosx] + ex[(-2a – 2b + 2a + 2bsinx]


= 0 = RHS.


Therefore, the given function is the solution of the corresponding differential equation.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz on DIfferential CalculusFREE Class
Functional Equations - JEE with ease48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses