Question

P.I of (D2 – 2D + 1)y = coshx is​

Answer 1

$\frac{1}{8}(2x^{2} e^{x} + e^{-x})$ is the P.I. of the differential equation $(D^2 - 2D + 1)y = coshx$

Step-by-step explanation:

Given: the differential equation $(D^2 - 2D + 1)y = coshx$

To Find: Particular Integral (P.I.) of the given differential equation

Solution:

• Finding the Particular Integral (P.I.) of the given differential equation

We have the differential equation $(D^2 - 2D + 1)y = coshx$ such that we can write,

$\Rightarrow y = \frac{1}{(D^2 - 2D + 1)} coshx$

$\Rightarrow y = \frac{1}{(D^2 - 2D + 1)} \Big( \frac{e^x \ + \ e^{-x}}{2} \Big) \ \because cosh x = \Big( \frac{e^x \ + \ e^{-x}}{2} \Big)$

$\Rightarrow y = \frac{1}{2} \Big( \frac{e^x}{D^2 - 2D + 1} \Big) + \frac{1}{2} \Big( \frac{e^{-x}}{D^2 - 2D + 1} \Big) \ \cdots \cdots (1)$

Now, considering

$y_1 = \frac{1}{2} \Big( \frac{e^x}{D^2 - 2D + 1} \Big)$

Using the rule $\ \frac{1}{f(D)}e^{ax} = \frac{1}{f''(a)}x^{2} e^{ax} \ \text{if} \ f(a)=0 \ \& \ f'(a)=0$, we get

$\Rightarrow y_1 = \frac{1}{2} \Big( \frac{e^x}{D^2 - 2D + 1} \Big) = \frac{1}{2} \Big( \frac{x^{2} e^x}{2} \Big) \ \cdots \cdots (2)$

Also, let

$y_2 = \frac{1}{2} \Big( \frac{e^{-x}}{D^2 - 2D + 1} \Big)$

Using the rule $\ \frac{1}{f(D)}e^{ax} = \frac{1}{f(a)}e^{ax}$, we get

$\Rightarrow y_2 = \frac{1}{2} \Big( \frac{e^x}{(-1)^2 - 2(-1) + 1} \Big) = \frac{1}{2} \Big( \frac{e^{-x}}{4} \Big) \ \cdots \cdots (3)$

Substituting (2) and (3) in (1), we get,

$y = \frac{1}{8}(2x^{2} e^{x} + e^{-x})$

Hence, $\frac{1}{8}(2x^{2} e^{x} + e^{-x})$ is the P.I. of the differential equation $(D^2 - 2D + 1)y = coshx$