The solution of the differential equation dy/dx + 2y = ex is- |
y = e-2x - Ce-3x y = e-2x + Ce-3x y = e-2x + Ce3x y = ex /3 + Ce-2x |
y = ex /3 + Ce-2x |
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
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