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alenahelenash · Accepted Answer

To solve the given third-order linear differential equation

{y^{″}}^{'} - y^{'} = 4 e^{- x} + 3 e^{2 x}

with the initial conditions

x = 0

,

y = 0

,

y^{'} = - 1

, and

y^{″} = 2

, we can use the method of solving linear homogeneous differential equations with constant coefficients.
The characteristic equation corresponding to the differential equation is:

r^{3} - r = 0

Factoring out

r

, we have:

r (r^{2} - 1) = 0

This equation has three roots:

r_{1} = 0

,

r_{2} = 1

, and

r_{3} = - 1

.
Therefore, the general solution to the homogeneous equation is:

y_{h} (x) = c_{1} e^{0 x} + c_{2} e^{1 x} + c_{3} e^{- 1 x}

Simplifying, we have:

y_{h} (x) = c_{1} + c_{2} e^{x} + c_{3} e^{- x}

To find the particular solution, we need to consider the non-homogeneous term

4 e^{- x} + 3 e^{2 x}

.
For the particular solution, we assume a solution of the form

y_{p} (x) = A e^{- x} + B e^{2 x}

, where

A

and

B

are constants to be determined.
Taking derivatives of

y_{p} (x)

, we have:

{y_{p}}^{'} (x) = - A e^{- x} + 2 B e^{2 x}

{y_{p}}^{″} (x) = A e^{- x} + 4 B e^{2 x}

{y_{p}}^{″}^{'} (x) = - A e^{- x} + 8 B e^{2 x}

Substituting these derivatives back into the original differential equation, we have:

(- A e^{- x} + 8 B e^{2 x}) - (- A e^{- x} + 2 B e^{2 x}) = 4 e^{- x} + 3 e^{2 x}

Simplifying, we get:

6 B e^{2 x} = 4 e^{- x} + 3 e^{2 x}

To equate the coefficients of the exponential terms on both sides, we have the following system of equations:

- 2 A + 6 B = 4

8 B = 3

Solving this system of equations, we find:

A = - \frac{7}{6}

and

B = \frac{3}{8}

Therefore, the particular solution is:

y_{p} (x) = - \frac{7}{6} e^{- x} + \frac{3}{8} e^{2 x}

The complete solution is the sum of the homogeneous and particular solutions:

y (x) = y_{h} (x) + y_{p} (x)

y (x) = c_{1} + c_{2} e^{x} + c_{3} e^{- x} - \frac{7}{6} e^{- x} + \frac{3}{8} e^{2 x}

Now, we can use the initial conditions to determine the values of the constants

c_{1}

,

c_{2}

, and

c_{3}

.
Given that

x = 0

,

y = 0

,

y^{'} = - 1

, and

y^{″} = 2

, we substitute these values into the complete solution and its derivatives:

y (0) = c_{1} + c_{2} + c_{3} - \frac{7}{6} + \frac{3}{8} = 0

y^{'} (0) = c_{2} - c_{3} + \frac{7}{6} + \frac{3}{4} = - 1

y^{″} (0) = c_{2} + c_{3} + \frac{7}{6} + \frac{3}{2} = 2

Simplifying these equations, we have a system of linear equations:

c_{1} + c_{2} + c_{3} = \frac{25}{24}

c_{2} - c_{3} = - \frac{25}{24}

c_{2} + c_{3} = \frac{11}{6}

We can solve this system to find the values of

c_{1}

,

c_{2}

, and

c_{3}

. Adding the second and third equations, we get:

2 c_{2} = \frac{11}{6} - \frac{25}{24}

Simplifying further:

2 c_{2} = \frac{22}{24} - \frac{25}{24}

2 c_{2} = - \frac{3}{24}

2 c_{2} = - \frac{1}{8}

c_{2} = - \frac{1}{16}

Substituting this value back into the second equation, we find:

- \frac{1}{16} - c_{3} = - \frac{25}{24}

c_{3} = - \frac{25}{24} + \frac{1}{16}

c_{3} = - \frac{25}{24} + \frac{3}{48}

c_{3} = - \frac{25}{24} + \frac{1}{16}

c_{3} = - \frac{50}{48} + \frac{3}{48}

c_{3} = - \frac{47}{48}

Finally, substituting the values of

c_{2}

and

c_{3}

back into the first equation, we have:

c_{1} - \frac{1}{16} - \frac{47}{48} = \frac{25}{24}

c_{1} = \frac{25}{24} + \frac{1}{16} + \frac{47}{48}

c_{1} = \frac{100}{96} + \frac{6}{96} + \frac{94}{96}

c_{1} = \frac{200}{96}

c_{1} = \frac{25}{12}

Therefore, the solution to the given differential equation with the specified initial conditions is:

y (x) = \frac{25}{12} + (- \frac{1}{16}) e^{x} - (\frac{47}{48}) e^{- x} - \frac{7}{6} e^{- x} + \frac{3}{8} e^{2 x}

Solve y'''-y'=4e^(-x)+3e^(2x) with the conditions x=0, y=0, y'=-1 and y"=2

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