Consider the differential equation: 2y′′−13y′−7y = 0

a. Show that, for any constants A and B, the following is a solution to the above differential equation: y = Ae^(−9x)+Be^(x/3)

b. Find the values A and B that make the above general solution into a solution for the following initial value problem: 2y′′−13y′−7y = 0; y(0) = 3, y′(0) = −5

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Mathematics

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Consider the differential equation: 2y′′−13y′−7y = 0

a. Show that, for any constants A and B, the following is a solution to the above differential equation: y = Ae^(−9x)+Be^(x/3)

b. Find the values A and B that make the above general solution into a solution for the following initial value problem: 2y′′−13y′−7y = 0; y(0) = 3, y′(0) = −5

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Joaobezerraid Joaobezerraid · Answer 1

a) Finding the derivatives and replacing into the equation, we reach an identity, an thus, for any value of the constants, [tex]y = Ae^{-7x} + Be^{\frac{x}{2}}[/tex] is a solution.
b) Solving a system of equations, according to the conditions, we find that: [tex]A = -\frac{7}{15}, B = \frac{52}{15}[/tex]

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Just a correction, the characteristic roots of the equation are [tex]y = 7[/tex] and [tex]y = -\frac{1}{2}[/tex], thus, we should test for:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

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Question a:

First, we find the derivatives, thus:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

[tex]y^{\prime} = 7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}[/tex]

[tex]y^{\prime\prime} = 49Ae^{-7x} + \frac{1}{4}Be^{-\frac{x}{2}}[/tex]

Now, we replace into the equation:

[tex]2y^{\prime\prime} - 13y^{\prime} - 7y = 0[/tex]

[tex]2(49Ae^{-7x} + \frac{1}{4}Be^{-\frac{x}{2}}) - 13(7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}) - 7(Ae^{7x} + Be^{-\frac{x}{2}}) = 0[/tex]

[tex]98Ae^{-7x} + \frac{1}{2}Be^{\frac{x}{2}} - 91Ae^{-7x} + \frac{13}{2}e^{-\frac{x}{2}} - 7Ae^{7x} - 7Be^{-\frac{x}{2}} = 0[/tex]

[tex]98Ae^{-7x} - 91Ae^{-7x} - 7Be^{-\frac{x}{2}} + \frac{1}{2}Be^{\frac{x}{2}} + \frac{13}{2}e^{-\frac{x}{2}} - 7Be^{-\frac{x}{2}} = 0[/tex]

[tex]0A + 0B = 0[/tex]

[tex]0 = 0[/tex], thus, we found the identity, and for each constant A and B, the following is a solution.

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Question b:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

Since [tex]y(0) = 3[/tex]

[tex]A + B = 3 \rightarrow B = 3 - A[/tex]

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[tex]y^{\prime} = 7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}[/tex]

Since [tex]y^{\prime}(0) = -5[/tex]

[tex]7A - \frac{1}{2}B = -5[/tex]

Using [tex]B = 3 - A[/tex]

[tex]7A - \frac{3}{2} + \frac{A}{2} = -5[/tex]

[tex]\frac{14A}{2} + \frac{A}{2} = -\frac{10}{2} + \frac{3}{2}[/tex]

[tex]\frac{15A}{2} = -\frac{7}{2}[/tex]

[tex]15A = -7[/tex]

[tex]A = -\frac{7}{15}[/tex]

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Then, B is given by:

[tex]B = 3 - A = 3 - (-\frac{7}{15}) = \frac{45}{15} + \frac{7}{15} = \frac{52}{15}[/tex]

Thus, the values are: [tex]A = -\frac{7}{15}, B = \frac{52}{15}[/tex]

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