Query_17

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course Mth 279

5/5

Query 17 Differential Equations*********************************************

Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

Our characteristic equation is

25 r^2 + 20 r + 4 = 0

with repeated solution r = -2/5

yielding solution set

{e^(-2/5 t), t e^(-2/5 t) }

and general solution

y(t) = A e^(-2/5 t) + B t e^(-2/5 t)

satisfying

y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

These conditions lead to the equations

A e^-2 + 5 B e^-2 = 4 e^-2

-2/5 A e^(-2) - 2/5 * 5 B e^-2 + B e^-2 = -3/5 e^-2,

or dividing both equations through by e^-2, and multiplying through the second by 5

A + 5 B = 4

-2 A - 5 B = -3.

The solution is A = 1, B = 3/5 ...

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Your solution:

Working through this problem the above solution seems to hold true,

and would be a bit redundant to type again.

y(t) = e^(-2t/5) + (3t/5)e^(-2t/5)

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1

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Question: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

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Your solution:

Using the Quadratic Equation:

r_1,2 = (-2*sqrt(3)) +,- sqrt((2*sqrt(3))^2 -4(3)(1)))/2(3)

r1 = (-2*sqrt(3))/12

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-2 sqrt(3) / (2 * 3) = -sqrt(3) / 3.

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and r2 = r1

General solution:

y(t) = Ae^(-2*sqrt(3)/12 t) + Bte^(-2*sqrt(3)/12 t)

y'(t) = A(-2*sqrt(3)/12)e^(-2*sqrt(3)/12 t) + B(1 - (-2*sqrt(3)/12 t))e^(-2*sqrt(3)/12 t)

y(0) = Ae^0 +B(0)e^0 = 2*sqrt(3)

Ae^0 = 2*sqrt(3)

A = 2*sqrt(3)

y'(0) = A(-2*sqrt(3)/12)e^0 + B(1 - 0)e^0 = 3

y'(0) = A(-2*sqrt(3)/12) + B = 3

y'(0) = (2*sqrt(3))(-2*sqrt(3)/12) + B = 3

1 + B = 3

B = 2

y(t) = (2*sqrt(3))e^(-2*sqrt(3)/12 t) + 2te^(-2*sqrt(3)/12 t)

@&

Good, except for the early arithmetic error. Compare with the following solution:

The characteristic equation is

3 r^2 + 2 sqrt(3) r + 1 = 0

with repeated solution

r = (-2 sqrt(3) +- sqrt( 12 - 12 ) ) / 6 = -sqrt(3)/3.

This yields fundamental set

{e^(-sqrt(3) / 3 * t), t e^(-sqrt(3) / 3 * t) }

with general solution

y(t) = A e^(-sqrt(3) / 3 * t) + B t e^(-sqrt(3) / 3 * t).

y(0) = A = 2 sqrt(3)

y ' (0) = -sqrt(3) / 3 * A + B = 3.

Substituting A = 2 sqrt(3) into the second equation we get

-6 + B = 3

so that

B = 9.

Our solution is therefore

y(t) = 2 sqrt(3) e^(-sqrt(3) / 3 * t) + 9 t e^(-sqrt(3) / 3 * t).

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2

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Given Solution:

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Question: Solve the equation

y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0,

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero.

See your text for a more complete statement of this problem.

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Your solution:

y2(t) = y1(t)u(t) = sin(t)u(t)

y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0

(sin(t)u(t))'' - 2 cot(t) (sin(t)u(t))' + (1 + 2 cot^2 t)(sin(t)u(t)) = 0

y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = sin(t)u(t)

u'' = 0, u(t) = a1*t + a2

y2(t) = sin(t)(a1*t + a2) = a1*tsin(t) + a2*sin(t)

let a2 = 0

y2(t) = a1*tsin(t)

y(t) = Asin(t) + B*t*sin(t)

The wronskian confirms that y1, y2 are a fundamental set.

W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)

|f '(t) g'(t)|

W(t) = |sin(t) tsin(t) |

|cos(t) sin(t) + tcos(t) |

= Sin^2(t) + tsin(t)cos(t) - tsin(t)cos(t) = sin^2(t)

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3

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Given Solution:

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