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

This question was previously asked in

BPSC Asstt. Prof. ME Held on Nov 2015 (Advt. 22/2014)

Option 1 : y = 2ex + e2x - xex

**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.

∴ m^{2} - 3m + 2 = 0

∴ m = 1 and 2.

∴ y = C_{1}e^{x} + C_{2}e^{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.

∴ **C _{1} + C_{2} = 3 **...... (1)

When x = 0, dy/dx = 3.

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

∴ 3 = C_{1} + 2C_{2} - 1

∴ **C _{1} + 2C_{2} = 4 ** ........ (2)

Solving (1) and (2) gives C_{1} = 2 and C_{2} = 1.

∴ **y = 2ex + e2x - xex**

Junior Executive (ATC) Official Paper 1: Held on Nov 2018 - Shift 1

20573

120 Questions
120 Marks
120 Mins