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1. Solve the initial value problem y 00 − 6y 0 + 9y = 0 with y(0) = 0, y 0 (0) = 2. 2. Find the general solution of y 00 + 6y 0 + 13y = 0. 3. The techniques and theory we have developed in Chapter 3 are naturally extended to higher order LHCC differential equations. Extend the methodology of Chapter 3 to find the general solution of 2y 000 − 4y 00 − 2y 0 + 4y = 0. 4. y1 = t 2 and y2 = 1 t are solutions of the differential equation t 2 y 00 − 2y = 0 for t > 0. Use the Wronskian to determine if these two solutions can be used as a fundamental set of solutions for this differential equation. 5. Create your own second order LHCC differential equation such that as t → ∞, the solutions grow without bound for some choices of initial conditions, but the solutions approach zero for other choices of initial conditions. Please clearly state your differential equation AND its general solution. 6. Consider a general second order LHCC differential equation ay00 + by0 + cy = 0. a) Show that if a > 0, b > 0, and c > 0, then limt→∞ y = 0 for all solutions y of this differential equation. b) If a > 0, b > 0, but c = 0, find the general solution of the resulting differential equation. In addition, compute limt→∞ y for solutions of the differential equation. 7. Consider the non-homogeneous differential equation y 00 − 3y 0 − 4y = g(t). For each g(t) given below, state the appropriate choice of form for yp that you would use to implement the method of undetermined coefficients. You do NOT need to solve for the coefficients in your yp; you do NOT need to state the solution of the original differential equation. a) g(t) = −e −t + e −t cos(t) b) g(t) = 2te4t + e 3t 8. Use variation of parameters to find the general solution of the differential equation y 00+3y 0−10y = e kt where k is an arbitrary constant.
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Explanation & Answer

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The characteristics equation is 𝑎2 − 6𝑎 + 9 = 0, it has one repeated real root 𝑎 = 3. Because of this,
the general solution is 𝑦𝑔 (𝑡) = 𝑒 3𝑡 (𝑐1 + 𝑐2 𝑡). Find the constants using the initial values:
𝑦(0) = 𝑐1 = 0,
𝑦 ′ (𝑡) = 3𝑒 3𝑡 (𝑐1 + 𝑐2 𝑡) + 𝑐2 𝑒 3𝑡 , 𝑦 ′ (0) = 3𝑐1 + 𝑐2 = 2,
so
𝑐1 = 0, 𝑐2 = 2, 𝑦(𝑡) = 𝟐𝒕𝒆𝟑𝒕 .
Here the characteristics equation is 𝑎2 + 6𝑎 + 13 = 0 and its solutions are complex:
𝑎1,2 = −3 ± √9 − 13 = −3 ± 2𝑖.
Because of this, the general solution is 𝑦𝑔 (𝑡) = 𝒆−𝟑𝒕 (𝒄𝟏 𝐜𝐨𝐬 𝟐𝒕 + 𝒄𝟐 𝐬𝐢𝐧 𝟐𝒕).
First, reduce by 2: 𝑦 ′′′ − 2𝑦 ′′ − 𝑦 ′ + 2𝑦 = 0. Here the characteristics equation is 𝑎3 − 2𝑎2 − 𝑎 + 2 = 0,
which is equivalent to 𝑎2 (𝑎 − 2) − (𝑎 − 2) = (𝑎2 − 1)(𝑎 − 2) = 0, so the roots are −1, 1, 2.
Because of this, the general solution is 𝑦𝑔 (𝑡) = 𝒄𝟏 𝒆−𝒕 + 𝒄𝟐 𝒆𝒕 + 𝒄𝟑 𝒆𝟐𝒕 .
The Wronskian is 𝑊(𝑡) = 𝑦1 (𝑡)𝑦2′ (𝑡) − 𝑦2 (𝑡)𝑦1′ (𝑡) =
1
1
= 𝑡 2 ∙ (− 2 ) − ∙ (2𝑡) = −1 − 2 = −3,
𝑡
𝑡
which is never zero. Because of this, these two solutions are linearly independent. And because the
order of the solution is 2, they are sufficient to form a fundamental set of solutions.
2
2
[Check that they are solutions: 𝑡 2 𝑦1′′ − 2𝑦1 = 2𝑡 2 − 2𝑡 2 = 0, 𝑡 2 𝑦2′′ − 2𝑦2 = 𝑡 2 ∙ (𝑡 3 ) − 𝑡 = 0.]
We need our equation to have both positive exponent and negative exponent solution. Let the
positive number be 2 and the negative be −1, then the equation for them is
(𝑎 − 2)(𝑎 + 1) = 𝑎2 − 𝑎 − 2 = 0
and the corresponding LHCC is 𝒚′′ − 𝒚′ − 𝟐𝒚 = 𝟎.
Its general solution is 𝑦 = 𝒄𝟏 𝒆𝟐𝒕 + 𝒄𝟐 𝒆−𝒕 , and if 𝑐1 ≠ 0 it tends to infinity while if 𝑐1 = 0 it tends to zero.
The characteristics equation is 𝑎𝑝2 + 𝑏𝑝 + 𝑐 = 0, its discriminant is 𝐷 = 𝑏 2 − 4𝑎𝑐 may be negative,
zero or positive.
𝑏
√−𝐷
𝑡)
2𝑎
If it is negative, the general solution has the form 𝑦𝑔 = 𝑒 −2𝑎𝑡 (𝑐1 cos (
√−𝐷
𝑡)),
2𝑎
+ 𝑐2 sin (
and
𝑏
because −...
