Question

The particular integral of (D2 − 2D + 1) y = cosh2x

Answer 1

Answer:

ji

Step-by-step explanation:

Hu with your company's needs at Nahi krt

Answer 2

Particular Integral of $(D^2 - 2D + 1)y = cosh \ 2x$ is $y = \dfrac{1}{18}(9e^{2x}+e^{-2x})$

Step-by-step explanation:

Given: Differential equation $(D^2 - 2D + 1)y = cosh \ 2x$

To Find: Particular Integral of the given differential equation

Solution:

• Finding the Particular Integral of the given differential equation

We have the differential equation $(D^2 - 2D + 1)y = cosh \ 2x$

Therefore, the solution is;

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

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

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

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

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

$\Rightarrow y = \dfrac{1}{2} \Big( \dfrac{e^{2x}}{1}\Big) + \dfrac{1}{2} \Big( \dfrac{e^{-2x}}{9}\Big)$

$\Rightarrow y = \dfrac{e^{2x}}{2} + \dfrac{e^{-2x}}{18}$

$\Rightarrow y = \dfrac{1}{18} (9e^{2x}+e^{-2x})$

Hence, the Particular Integral is $y = \dfrac{1}{18}(9e^{2x}+e^{-2x})$