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

y =

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 = and m =

thus,

the general equation comes as:

y =

or

y =

now,

at y(0) = 1

therefore,

1 =

or

C₁ + C₂ = 1  .............(1)

and,

y' =

also,

y'(0) = 5

thus,

5 =

or

3C₂ - C₁ = 10 ...........(2)

on adding 1 and 2, we get

     C₁ + C₂ = 1

+ (- C₁ + 3C₂) = 10

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

4C₂ = 11

or

C₂ =

thus,

   C₁ + C₂ = 1

or

   C₁ +  = 1

or

C₁ =  

Hence,

The solution is y =

on substituting the values,

y =

you answer is a

step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°