The solution of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = {e^x},\;y = 3\) and \(\frac{{dy}}{{dx}} = 3\), when x = 0 is

Concept:

The solution of a linear differential equation of n^th order comprises of two parts. They are:

• Complementary function
• Particular integral

i.e. y = C.F. + P.I.

Calculation:

Given:

\(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = {e^x},\;y = 3\;and\;\frac{{dy}}{{dx}} = 3\;when\;x=0\)

Complimentary function (CF):

Put R.H.S = 0, and calculate the roots.

\(\frac{{{d^2}y}}{{d{x^2}}} - 3\frac{{dy}}{{dx}} + 2y = 0\)

For finding the roots put d/dx = m.

∴ m² - 3m + 2 = 0

∴ m = 1 and 2.

∴ y = C₁e^x + C₂e^2x

Particular Integral (PI):

\(\frac{1}{D^2\;-\;3D\;+\;2}e^x\)

\(\frac{1}{2D\;-\;3}xe^x\)    [∵ f(D) = 0 when D = 1]

∴ P.I = -xe^x

General solution is y = C1ex + C2e2x - xe^x

When x = 0, y = 3.

∴ C₁ + C₂ = 3 ...... (1)

When x = 0, dy/dx = 3.

\(\frac{dy}{dx}=C_1e^x\;+\;2C_2e^{2x}\;-\;xe^x\;-\;e^x\)

∴ 3 = C₁ + 2C₂ - 1

∴ C₁ + 2C₂ = 4  ........ (2)

Solving (1) and (2) gives C₁ = 2 and C₂ = 1.

∴ y = 2ex + e2x - xex