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Problem 25: Reduction of Order

Given the second-order linear differential equation:

\[t^2 y'' + 3ty' + y = 0, \quad t > 0, \quad y_1 = t^{-1}\]

We assume a second solution of the form \(y_2 = v y_1 = v t^{-1}\). Calculating the derivatives:

\[\begin{aligned} y_2 &= v t^{-1} \\ y_2' &= -v t^{-2} + v' t^{-1} \\ y_2'' &= 2v t^{-3} - v' t^{-2} - v' t^{-2} + v'' t^{-1} \\ &= 2v t^{-3} - 2v' t^{-2} + v'' t^{-1} \end{aligned}\]

Substituting these into the original differential equation:

\[2v t^{-1} - 2v' + v'' t - 3v t^{-1} + 3v' + v t^{-1} = 0\]

Simplifying the expression by canceling terms:

\[v'' t + v' = 0\]

Using the substitution \(v'' = \frac{d(v')}{dt}\):

\[\begin{aligned} t \frac{d(v')}{dt} &= -(v') \\ \frac{1}{(v')} d(v') &= -\frac{1}{t} dt \\ \ln(v') &= -\ln t + C \implies v' = C t^{-1} \\ v &= C_1 \ln t + C_2 \end{aligned}\]

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Problem 23: Reduction of Order

Given the second-order linear differential equation:

\[t^2 y'' - 4ty' + 6y = 0, \quad t > 0, \quad y_1 = t^2\]

We assume a second solution of the form \(y_2 = v t^2\). Calculating the derivatives:

\[\begin{aligned} y_2 &= v t^2 \\ y_2' &= 2vt + v' t^2 \\ y_2'' &= v'' t^2 + 4v' t + 2v \end{aligned}\]

Substituting these into the original differential equation:

\[v'' t^4 + 4v' t^3 + 2v t^2 - 8v t^2 - 4v' t^3 + 6v t^2 = 0\]

After canceling the terms, we are left with:

\[v'' t^4 = 0 \implies v'' = 0 \quad \text{(since } t \neq 0 \text{ in general)}\]

Integrating twice to find \(v\):

\[\begin{aligned} v' &= C_1 \\ v &= C_1 t + C_2 \end{aligned}\]

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Problem 15, Part (c)

From part a), we have the equation:

\[ y = e^{-3/2 t} - \frac{5}{2} t e^{-3/2 t} \]

Find \( (t_0, y_0) \) of the minimum point.

To find the minimum, set the first derivative to zero:

\[ y' = 0 \]

\[ y' = \frac{1}{4} e^{-3t/2} (15t - 16) \]

Setting \( y' = 0 \) leads to:

\[ y' = 0 \rightarrow t = 16/15 \]

Substituting back to find the y-coordinate:

\[ y = -\frac{5}{3} e^{-8/5} \]

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3.5 Nonhomogeneous Eqs: Method of Undetermined Coefficients

Linear, constant-coefficient, but not homogeneous differential equations take the form:

\[ y'' + ay' + by = g(t) \]

Note on \( g(t) \):

Undetermined coefficients can handle: polynomial, \( \sin \) or/and \( \cos \), and exponentials.

Example

\[ y'' - y' - 2y = e^{3t} \]

Solution:

\[ y = y_h + Y \]

\( \uparrow \) homogeneous part (\( y_h \))
\( \nearrow \) particular solution due to \( g(t) \) (\( Y \))

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here, \( y_h = C_1 e^{-t} + C_2 e^{2t} \)

\[ y = C_1 e^{-t} + C_2 e^{2t} + Y \]

particular solution \( Y \) will resemble \( g(t) \)

here, \( g(t) = e^{3t} \), assume \( Y = A e^{3t} \)

undetermined coeff.

form from \( g(t) \)

\( Y \) is a solution, so satisfies DE

\[ Y = A e^{3t} \quad Y' = 3A e^{3t} \quad Y'' = 9A e^{3t} \]

DE \( y'' - y' - 2y = e^{3t} \)

\[ 9A e^{3t} - 3A e^{3t} - 2A e^{3t} = e^{3t} \rightarrow A = 1/4 \]

\[ y = C_1 e^{-t} + C_2 e^{2t} + \frac{1}{4} e^{3t} \]

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example

\( y'' - y' - 2y = e^{2t} \) same LHS as prev. example

\[ y = C_1 e^{-t} + C_2 e^{2t} + Y \]

following last example, assume \( Y = A e^{2t} \)

but note this form duplicates part of homogeneous solution (similar to repeated roots)

modify \( Y \): \( Y = A t e^{2t} \)

\[ Y' = 2A t e^{2t} + A e^{2t} \]\[ Y'' = 4A t e^{2t} + 4A e^{2t} \]

sub into DE

\[ 4A t e^{2t} + 4A e^{2t} - 2A t e^{2t} - A e^{2t} - 2A t e^{2t} = e^{2t} \]\[ 3A e^{2t} = e^{2t} \quad A = 1/3 \]

\[ y = C_1 e^{-t} + C_2 e^{2t} + \frac{1}{3} t e^{2t} \]

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Example: Method of Undetermined Coefficients

Original equation (with correction):

\[ y'' - 4y = e^{2t} \]

The general solution is given by the sum of the homogeneous solution and the particular solution:

\[ y = c_1 e^{2t} + c_2 t e^{2t} + Y \]

For the particular solution \( Y \), we assume a form based on the non-homogeneous term. However, since \( e^{2t} \) and \( t e^{2t} \) are already part of the homogeneous solution, we must multiply by \( t \) until there is no duplication.

\[ Y = A e^{2t} t^2 \]

Keep multiplying by \( t \) until not duplicating ANY part of homo. solution.

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Example: Trigonometric Non-homogeneous Term

\[ y'' - y' - 2y = \cos t \]

Same LHS as example 1

The general solution is:

\[ y = c_1 e^{-t} + c_2 e^{2t} + Y \]

If \( g(t) \) is \( \sin(\alpha t) \) or \( \cos(\alpha t) \), assume:

\[ Y = A \cos(\alpha t) + B \sin(\alpha t) \]

\( \sin \), \( \cos \) always show up together in \( Y \), even if \( g(t) \) only has one of them.

Here, for \( g(t) = \cos t \):

\[ \begin{aligned} Y &= A \cos t + B \sin t \\ Y' &= -A \sin t + B \cos t \\ Y'' &= -A \cos t - B \sin t \end{aligned} \]

plug into DE

Substituting into the differential equation:

\[ -A \cos t - B \sin t + A \sin t - B \cos t - 2A \cos t - 2B \sin t = \cos t \]

Grouping terms:

\[ (-3A - B) \cos t + (A - 3B) \sin t = 1 \cos t + 0 \sin t \]

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\[ \begin{cases} -3A - B = 1 \\ A - 3B = 0 \end{cases} \]

\( A = -3/10, \quad B = -1/10 \)

\[ y = C_1 e^{-t} + C_2 e^{2t} - \frac{3}{10} \cos t - \frac{1}{10} \sin t \]

Example

\( y'' + y = \sin t \)

\( r^2 + 1 = 0 \implies r = \pm i \)

\( y = C_1 \cos t + C_2 \sin t + Y \)

\( Y = (A \cos t + B \sin t)t \)

to avoid duplicating homo. solution

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Example

\( y'' - y' - 2y = 1 \)

LHS same as first example

RHS polynomial

\( y = C_1 e^{-t} + C_2 e^{2t} + Y \)

\( Y \) has form of polynomial of the same degree as \( g(t) \)

\( Y = A \) (0th degree)

Example

\( y'' - y' - 2y = t^3 \)

\( y = C_1 e^{-t} + C_2 e^{2t} + Y \)

\( Y = At^3 + Bt^2 + Ct + D \) (3rd deg.)

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Differential Equations: Method of Undetermined Coefficients

Example 1

\[ y'' - y' = t^2 \]

The general solution is given by:

\[ y = C_1 + C_2 e^t + Y \]

For the particular solution \( Y \), we initially consider a second-degree polynomial. However, we must multiply by \( t \) to avoid duplication with the homogeneous solution:

\[ Y = (At^2 + Bt + C)t \]

Note: We multiply by \( t \) because the constant term \( C \) duplicates the constant \( C_1 \) in the homogeneous solution. The term in parentheses is a 2nd degree polynomial.

Example 2

\[ y'' + y = t \sin t \]

The homogeneous solution is:

\[ y = C_1 \cos t + C_2 \sin t + Y \]

Analyzing the non-homogeneous term \( t \sin t \):

\( t \) is a first degree polynomial.
The trigonometric part can be \( \sin t \) or \( \cos t \).

To eliminate duplication with the homogeneous solution, we multiply the standard form by \( t \):

\[ Y = t(At + B) \sin t + t(Ct + D) \cos t \]

The extra factor of \( t \) is included to eliminate duplication with the terms in the homogeneous solution.