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Exercise 0.2.101: Show that x = e−2tis a solution to x′′+ 4x′+ 4x = 0.

Exercise 0.2.102: Is y = x2a solution to x2y′′− 2y = 0? Justify.
Exercise 0.2.103:Letxy′′− y′= 0. Try a solution of the formy = xr. Is this a solution for somer?
If so, find all such r.
Exercise 0.2.104:Verify thatx = C1et+ C2is a solution tox′′− x′= 0. FindC1andC2so thatx
satisfies x(0) = 10 and x′(0) = 100.
dϕ
Exercise 0.2.105: Solve
= 8ϕ and ϕ(0) = −9.
ds

1 answer

Let's tackle each exercise one by one:

Exercise 0.2.101

We want to show that \( x = e^{-2t} \) is a solution to the differential equation \( x'' + 4x' + 4x = 0 \).

  1. First, calculate the first and second derivatives of \( x \):

    • \( x' = -2e^{-2t} \)
    • \( x'' = 4e^{-2t} \)

  2. Substitute \( x \), \( x' \), and \( x'' \) into the differential equation: \[ x'' + 4x' + 4x = 4e^{-2t} + 4(-2e^{-2t}) + 4e^{-2t} \] Simplifying: \[ = 4e^{-2t} - 8e^{-2t} + 4e^{-2t} = 0 \] Since the left-hand side equals zero, \( x = e^{-2t} \) is indeed a solution.

Exercise 0.2.102

We need to check if \( y = x^2 \) is a solution to the equation \( x^2 y'' - 2y = 0 \).

  1. Calculate the first and second derivatives of \( y \):

    • \( y' = 2x \)
    • \( y'' = 2 \)

  2. Substitute \( y \), \( y' \), and \( y'' \) into the differential equation: \[ x^2 (2) - 2(x^2) = 2x^2 - 2x^2 = 0 \] This is true, so \( y = x^2 \) is indeed a solution to the equation.

Exercise 0.2.103

We have the equation \( xy'' - y' = 0 \) and we want to try a solution of the form \( y = x^r \).

  1. Calculate the first and second derivatives:

    • \( y' = r x^{r-1} \)
    • \( y'' = r(r-1)x^{r-2} \)

  2. Substitute into the equation: \[ x \cdot (r(r-1)x^{r-2}) - (r x^{r-1}) = r(r-1)x^{r-1} - r x^{r-1} = (r(r-1) - r)x^{r-1} = (r^2 - 2r)x^{r-1} \] Set this equal to zero: \[ (r^2 - 2r)x^{r-1} = 0 \] For this to hold for all \( x \), we need \( r^2 - 2r = 0 \).

  3. Solving the quadratic: \[ r(r - 2) = 0 \implies r = 0 \text{ or } r = 2 \] Therefore, solutions exist for \( r = 0 \) and \( r = 2 \).

Exercise 0.2.104

We want to verify that \( x = C_1 e^t + C_2 \) is a solution to \( x'' - x' = 0 \), and find \( C_1 \) and \( C_2 \) such that \( x(0) = 10 \) and \( x'(0) = 100 \).

  1. Calculate derivatives:

    • \( x' = C_1 e^t \)
    • \( x'' = C_1 e^t \)

  2. Substitute into the differential equation: \[ C_1 e^t - C_1 e^t = 0 \] This shows that \( x = C_1 e^t + C_2 \) is indeed a solution.

  3. Now use the initial conditions:

    • At \( t = 0 \): \[ x(0) = C_1 + C_2 = 10 \]
    • The derivative at \( t = 0 \): \[ x'(0) = C_1 e^0 = C_1 = 100 \]

  4. From \( C_1 = 100 \): \[ 100 + C_2 = 10 \implies C_2 = 10 - 100 = -90 \] Thus, \( C_1 = 100 \) and \( C_2 = -90 \).

Exercise 0.2.105

We need to solve the equation \( \frac{d\varphi}{ds} = 8\varphi \) with the condition \( \varphi(0) = -9 \).

  1. This is a first-order linear ordinary differential equation. We can solve it using separation of variables: \[ \frac{d\varphi}{\varphi} = 8 ds \]

  2. Integrating both sides: \[ \ln |\varphi| = 8s + C \]

  3. Exponentiating, we have: \[ \varphi = e^{8s + C} = e^Ce^{8s} \] Let \( A = e^C \): \[ \varphi = A e^{8s} \]

  4. Using the initial condition \( \varphi(0) = -9 \): \[ -9 = A e^{0} \implies A = -9 \]

  5. Therefore: \[ \varphi(s) = -9 e^{8s} \]

In summary, the solutions are:

  • Exercise 0.2.101: \( x = e^{-2t} \) is a solution.
  • Exercise 0.2.102: \( y = x^2 \) is a solution.
  • Exercise 0.2.103: Solutions exist for \( r = 0 \) and \( r = 2 \).
  • Exercise 0.2.104: \( C_1 = 100 \), \( C_2 = -90 \).
  • Exercise 0.2.105: \( \varphi(s) = -9 e^{8s} \).