Question

Find the differential equation whose solution is y= A sin x +B Cos X.​

Answer 1

Question :

Find the differential equation whose solution is y = AsinX + BcosX.

Solution:

The given equation is ;

y = AsinX + BcosX ---------(1)

Now ,

Differentiating the eq-(1) both sides with respect to X , we get;

=> dy/dx = d(AsinX)/dx + d(BcosX)/dx

=> y' = A•d(sinX)/dx + B•d(cosX)/dx

=> y' = AcosX + B(-sinX)

=> y' = AcosX - BsinX ----------(2)

Again;

Differentiating the eq-(2) both sides with respect to X , we get;

=> dy'/dx = d(AcosX)/dx - d(BsinX)/dx

=> y" = A•d(cosX)/dx - B•d(sinX)/dx

=> y" = A(- sinX) - BcosX

=> y" = - AsinX - BcosX

=> y" = - ( AsinX + BcosX )

=> y" = - y {using eq-(1)}

=> y" + y = 0

Hence,

The required differential equation is ;

y" + y = 0.

Note:

• y' denotes first derivative of y.
• y" denotes second derivative of y.
• If y = sinX , then y' = cosX
• If y = cosX , then y' = - sinX

Answer 2

Answer:

y=,a sin x+b sin x,then find the differential equation whose solution is y