Let f(x) be a polynomial satisfying f(0) = 2, f'(0) = 3 and f''(x) = f

1.	Let f(x) be a polynomial satisfying f(0) = 2, f'(0) = 3 and f''(x) = f(x). Then f(4) is equal to1. \(5 \frac {(e^8 - 1)}{2e^4}\)2. \(\frac {(5e^8 - 1)}{2e^4}\)3. \(\frac {2e^4}{5e^8 - 1}\)4. \(\frac {2e^4}{5(e^8 + 1)}\)
Answer» Correct Answer - Option 2 : \(\frac {(5e^8 - 1)}{2e^4}\) CONCEPT : If roots are real and different then complimentary solution is given by F(x) = c1 em x + c2 en x CALCULATION : Differential equation f"(x) = f(x) f"(x) - f(x) = 0 [D2-1] f(x) = 0 The corresponding auxiliary equation D2 - 1 = 0 Root of auxiliary equation D2 = 1 ⇒ D = \(±\)1 Here, m = 1 and n = - 1 Roots are real and different. So complimentary solution is given by ⇒ f(x) = c1 ex + c2 e-x Put initial condition f(0) = 2 in the above equation we get, ⇒ 2 = c1 e0 + c2 e0 ⇒ 2 = c1 + c2 -----(1) ∵ f(x) = c1 ex + c2 e-x ⇒ f`(x) = c1 ex - c2 e-x Put initial condition f`(0) = 3 in the above equation ⇒ 3 = c1 - c2 -----(2) By adding equation (1) and equation (2) we get, ⇒ 2c1 =5 ⇒ c1 = 5/2 Put value of c₁ in equation (1) ⇒ 2 = 5/2 + c2 ⇒c 2 = - 1/2 Solution of the given differential equation f(x) = \(\frac{5}{2}\)ex - \(\frac{1}{2}\)e-x \(\Rightarrow f(x) = \frac {(5e^{2x} - 1)}{2e^x}\) So, the value of f(4) \(\Rightarrow f(4) = \frac {(5e^8 - 1)}{2e^4}\) Hence, option B is the correct answer.

Let f(x) be a polynomial satisfying f(0) = 2, f'(0) = 3 and f''(x) = f(x). Then f(4) is equal to1. \(5 \frac {(e^8 - 1)}{2e^4}\)2. \(\frac {(5e^8 - 1)}{2e^4}\)3. \(\frac {2e^4}{5e^8 - 1}\)4. \(\frac {2e^4}{5(e^8 + 1)}\)

Answer» Correct Answer - Option 2 : \(\frac {(5e^8 - 1)}{2e^4}\)

CONCEPT :

If roots are real and different then complimentary solution is given by F(x) = c1 em x + c2 en x

CALCULATION :

Differential equation f"(x) = f(x)

f"(x) - f(x) = 0

[D2-1] f(x) = 0

The corresponding auxiliary equation D2 - 1 = 0

Root of auxiliary equation D2 = 1

⇒ D = \(±\)1

Here, m = 1 and n = - 1

Roots are real and different. So complimentary solution is given by

⇒ f(x) = c1 ex + c2 e-x

Put initial condition f(0) = 2 in the above equation we get,

⇒ 2 = c1 e0 + c2 e0

⇒ 2 = c1 + c2 -----(1)

∵ f(x) = c1 ex + c2 e-x

⇒ f`(x) = c1 ex - c2 e-x

Put initial condition f`(0) = 3 in the above equation

⇒ 3 = c1 - c2 -----(2)

By adding equation (1) and equation (2) we get,

⇒ 2c1 =5 ⇒ c1 = 5/2

Put value of c₁ in equation (1)

⇒ 2 = 5/2 + c2 ⇒c 2 = - 1/2

Solution of the given differential equation f(x) = \(\frac{5}{2}\)ex - \(\frac{1}{2}\)e-x

\(\Rightarrow f(x) = \frac {(5e^{2x} - 1)}{2e^x}\)

So, the value of f(4)

\(\Rightarrow f(4) = \frac {(5e^8 - 1)}{2e^4}\)

Hence, option B is the correct answer.