Answer:
a)<em> y₂ (x) = e ⁵ˣ </em>
<em>Complementary function</em>
![y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }](https://tex.z-dn.net/?f=y_%7BC%7D%20%3D%20C_%7B1%7D%20%7Be%5E%7B-5x%7D%20%7D%20%2B%20C_%7B2%7D%20%7Be%5E%7B5x%7D%20%7D)
<em>b) particular integral</em>
![P.I = y_{p} = \frac{-4}{25}](https://tex.z-dn.net/?f=P.I%20%3D%20y_%7Bp%7D%20%3D%20%5Cfrac%7B-4%7D%7B25%7D)
<em></em>
Step-by-step explanation:
<u><em>step(i):</em></u>-
<em>Given differential equation y''-25y= 4</em>
<em>operator form </em>
<em> ⇒ D²y - 25 y =4</em>
⇒ (D² - 25) y =4
This is the form of f(D)y = ∝(x)
where f(m) = D² - 25 and ∝(x) =4
<em>The auxiliary equation A(m) =0</em>
<em> ⇒ m² - 25 =0</em>
m² - 5² =0
⇒ (m+5)(m-5) =0
<em> ⇒ m =-5 , 5</em>
<em>Complementary function</em>
![y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }](https://tex.z-dn.net/?f=y_%7BC%7D%20%3D%20C_%7B1%7D%20%7Be%5E%7B-5x%7D%20%7D%20%2B%20C_%7B2%7D%20%7Be%5E%7B5x%7D%20%7D)
This is form of
![y_{C} = C_{1} y_{1} (x) + C_{2} y_{2} (x)](https://tex.z-dn.net/?f=y_%7BC%7D%20%3D%20C_%7B1%7D%20y_%7B1%7D%20%28x%29%20%2B%20C_%7B2%7D%20y_%7B2%7D%20%28x%29)
<em>where y₁ (x) = e⁻⁵ˣ and y₂ (x) = e ⁵ˣ </em>
<u><em>Step(ii):</em></u><em>-</em>
<u><em>Particular integral:-</em></u>
<u><em> </em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
<em> = </em>
<em></em>
<em> put D = 0</em>
The particular integral
![y_{p} = \frac{1}{ -25} 4](https://tex.z-dn.net/?f=y_%7Bp%7D%20%3D%20%5Cfrac%7B1%7D%7B%20-25%7D%20%204)
![P.I = y_{p} = \frac{-4}{25}](https://tex.z-dn.net/?f=P.I%20%3D%20y_%7Bp%7D%20%3D%20%5Cfrac%7B-4%7D%7B25%7D)
<u><em>Conclusion:-</em></u>
General solution of given differential equation
![y = y_{C} +y_{P}](https://tex.z-dn.net/?f=y%20%3D%20y_%7BC%7D%20%2By_%7BP%7D)
![y = C_{1} {e^{-5x} } + C_{2} {e^{5x} } -\frac{4}{25}](https://tex.z-dn.net/?f=y%20%3D%20C_%7B1%7D%20%7Be%5E%7B-5x%7D%20%7D%20%2B%20C_%7B2%7D%20%7Be%5E%7B5x%7D%20%7D%20-%5Cfrac%7B4%7D%7B25%7D)