Question

5. Solve the given initial value problem 4y" – 4y' – 3y = 0, y(0) = 1, y(0) = 5

Answer 1

Answer:

y = $\frac{-7}{4}e^{\frac{-1}{2}x}+ \frac{11}{4}e^{\frac{3}{2}x}$

Step-by-step explanation:

Given:

4y" – 4y' – 3y = 0, y(0) = 1, y'(0) = 5

Now,

The auxiliary equation will be given as:

4m² - 4m - 3 = 0

or

4m² + 2m - 6m - 3 = 0

or

2m (2 + 1) - 3 (2m + 1) = 0

or

( 2m - 3 ) × ( 2m + 1 ) = 0

therefore,

m = $\frac{\textup{3}}{\textup{2}}$ and m = $\frac{\textup{-1}}{\textup{2}}$

thus,

the general equation comes as:

y = $C_1e^{mx}+ C_2e^{mx}$

or

y = $C_1e^{\frac{-1}{2}x}+ C_2e^{\frac{3}{2}x}$

now,

at y(0) = 1

therefore,

1 = $C_1e^{\frac{-1}{2}\times0}+ C_2e^{\frac{3}{2}\times0}$

or

C₁ + C₂ = 1  .............(1)

and,

y' = $\frac{-1}{2}C_1e^{\frac{-1}{2}x}+ \frac{3}{2}C_2e^{\frac{3}{2}x}$

also,

y'(0) = 5

thus,

5 = $\frac{-1}{2}C_1e^{\frac{-1}{2}\times0}+ \frac{3}{2}C_2e^{\frac{3}{2}\times0}$

or

3C₂ - C₁ = 10 ...........(2)

on adding 1 and 2, we get

     C₁ + C₂ = 1

+ (- C₁ + 3C₂) = 10

===============

4C₂ = 11

or

C₂ = $\frac{\textup{11}}{\textup{4}}$

thus,

   C₁ + C₂ = 1

or

   C₁ + $\frac{\textup{11}}{\textup{4}}$  = 1

or

C₁ =  $\frac{\textup{-7}}{\textup{4}}$

Hence,

The solution is y = $C_1e^{mx}+ C_2e^{mx}$

on substituting the values,

y = $\frac{-7}{4}e^{\frac{-1}{2}x}+ \frac{11}{4}e^{\frac{3}{2}x}$