Question

The particular integral of the differential equation (D2 + 4)y = sinh (x) is​

Answer 1

Answer:

sorry I didn't know that

Answer 2

The particular integral of the given differential equation is $\dfrac{1}{5} sinh \ x$

Step-by-step explanation:

Given: Differential equation $(D^{2} + 4 ) y = sinh \ (x)$

To Find: Particular integral of the given differential equation

Solution:  

• Finding the particular integral of the given differential equation

For the given differential equation, the particular integral can be determined by;

$y_{_{PI}} = \dfrac{1}{D^{2} + 4} sinh \ x$

$\Rightarrow y_{_{PI}} = \dfrac{1}{D^{2} + 4} \Big( \dfrac{e^x - e^{-x}}{2} \Big) \ \ \ \ \ \because \ sinh \ x = \Big( \dfrac{e^x - e^{-x}}{2} \Big)$

$\Rightarrow y_{_{PI}} = \dfrac{1}{2} \Big[ \dfrac{1}{D^{2} + 4} e^x - \dfrac{1}{D^{2} + 4} e^{-x} \Big]$

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

$\Rightarrow y_{_{PI}} = \dfrac{1}{2} \Big[ \dfrac{1}{(1)^{2} + 4} e^x - \dfrac{1}{(-1)^{2} + 4} e^{-x} \Big]$

$\Rightarrow y_{_{PI}} = \dfrac{1}{2} \Big[ \dfrac{1}{5} e^x - \dfrac{1}{5} e^{-x} \Big]$

$\Rightarrow y_{_{PI}} = \dfrac{1}{5} \Big[ \dfrac{e^x - e^{-x}}{2} \Big]$

$\Rightarrow y_{_{PI}} = \dfrac{1}{5} sinh \ x$

Hence, the particular integral of the given differential equation is $\dfrac{1}{5} sinh \ x$