Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

The solution of the differential equation dy/dx + 2y = ex is-

Options:

y = e-2x - Ce-3x

y = e-2x + Ce-3x

y = e-2x + Ce3x

y = ex /3 + Ce-2x

Correct Answer:

y = ex /3 + Ce-2x

Explanation:

The given differential equation is dy/dx + 2y = ex

This ifs of the form dy/dx +Py = Q

where P = 2, Q= ex

I.F. = e∫Pdx = e∫2dx =e2x 

The solution of the given differential equation is given by-

y(I.F.) = ∫(Q xI.F.)dx + C

⇒ye2x = ∫(ex.e2x)dx + C 

⇒ ye2x = ∫(e3x)dx + C 

⇒ye2x = e3x/3 + C

so, y = ex /3 + Ce-2x