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PAGE 1

Differential Equations: Problem 32

\[ 2y' + ty = 2 \]

From equation 47:

\[ y = e^{-t^2/4} \int_{0}^{t} e^{s^2/4} \, ds + ce^{-t^2/4} \]

We want to find the limit as \( t \to \infty \):

\[ \lim_{t \to \infty} y = ? \]

\[ y = \frac{\int_{0}^{t} e^{s^2/4} \, ds}{e^{t^2/4}} \]

Figure: A rough sketch of a graph with axes and a curve representing the integral function.

Let \( f(t) = \int_{0}^{t} e^{s^2/4} \, ds \) and \( g(t) = e^{t^2/4} \).

As \( t \to \infty \), we have an indeterminate form:

\[ \lim_{t \to \infty} y = \frac{\infty}{\infty} \]

Use l'Hospital's Rule (derivative of top & bottom):

\[ = \lim_{t \to \infty} \frac{e^{t^2/4}}{\frac{t}{2} e^{t^2/4}} = 0 \]

Note on the integral growth:

\[ \int_{0}^{t} s^2 \, ds = \left[ \frac{1}{3} s^3 \right]_0^t = \frac{1}{3} t^3 \]

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Differential Equations: Problem 24

\[ ty' + (t+1)y = 2te^{-t}, \quad y(1) = a, \, t > 0 \]

Solving for \( y \):

\[ y = te^{-t} + ce^{-t}/t \]

Using the initial condition \( y(1) = a \):

\[ a = e^{-1} + ce^{-1} = e^{-1}(1+c) \]

\[ ae = 1 + c \implies c = ae - 1 \]

Substituting \( c \) back into the general solution:

\[ y = \frac{t}{e^t} + (ae - 1) \frac{1}{te^t} \]

As \( t \to 0 \), the first term \( \frac{t}{e^t} \to 0 \).
As \( t \to 0 \), the second term \( \frac{1}{te^t} \to \infty \).

Behavior as \( t \to 0 \):

\( y \to \infty \)
\( y \to -\infty \)
\( y \to 0 \) (Target behavior)

What is the value of \( a \)?

We want \( y \to 0 \), so choose \( a \) so that the coefficient of the divergent term is zero:

\[ ae - 1 = 0 \]

\[ a = \frac{1}{e} \]

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2.2 Separable Equations

Hw #28 is optional

look at

\[ y' = \frac{x^2}{y(1+x^3)} \]

y: dependent variable
x: indep. variable

\[ \frac{dy}{dx} = \frac{x^2}{y(1+x^3)} \]

multiply \( y \, dx \) so each variable is on its own side

\[ y \, dy = \frac{x^2}{1+x^3} \, dx \]

integrate

\[ \int y \, dy = \int \frac{x^2}{1+x^3} \, dx \]

\[ u = 1 + x^3 \]\[ du = 3x^2 \, dx \]\[ dx = \frac{1}{3x^2} \, du \]

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\[ \frac{1}{2}y^2 = \int \frac{x^2}{u} \frac{1}{3x^2} \, du = \int \frac{1}{3u} \, du \]

\[ = \frac{1}{3} \ln |u| + C \]

\[ \frac{1}{2}y^2 = \frac{1}{3} \ln |1+x^3| + C \]

put +C w/ x

implicit form of solution

often written as

multiply by 6 to eliminate fractions

\[ 3y^2 = 2 \ln |1+x^3| + C \]

\( \leftarrow \) 6 times previous C

\[ 3y^2 - 2 \ln |1+x^3| = C \]

or

\[ y^2 = \frac{2}{3} \ln |1+x^3| + C \]

\[ y = \left( \frac{2}{3} \ln |1+x^3| + C \right)^{1/2} \]

explicit form

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Example: Solving a Separable Differential Equation

\[ y' = \frac{1-2x}{y}, \quad y(1) = -2 \]

\[ \frac{dy}{dx} = \frac{1-2x}{y} \]

separate by multiplication/div

NEVER by add/subtraction

\[ y \, dy = (1-2x) \, dx \]

\[ \int y \, dy = \int (1-2x) \, dx \]

\[ \frac{1}{2}y^2 = x - x^2 + C \]

Use Initial Condition (IC)

Given: \( y(1) = -2 \)

\[ \frac{1}{2}(-2)^2 = 1 - (1)^2 + C \]

so \( C = 2 \)

\[ \frac{1}{2}y^2 = x - x^2 + 2 \]

\[ y^2 = 2x - 2x^2 + 4 \]

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\[ y = \pm \sqrt{2x - 2x^2 + 4} \]

can have only one of these as a solution,

use IC: \( y(1) = -2 \), so here, choose negative

\[ y = -\sqrt{2x - 2x^2 + 4} \]

On what interval is this solution valid?

can't take square root of neg. #

so \( 2x - 2x^2 + 4 \ge 0 \)

Why / why not?

NO, because \( y = 0 \) breaks the DE

\[ y' = \frac{1-2x}{y} \]

even though \( y = -\sqrt{2x - 2x^2 + 4} \) is still defined

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Interval of Validity Analysis

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Solving the Inequality

\[ \begin{aligned} x - x^2 + 2 &> 0 \\ -(x^2 - x - 2) &> 0 \\ (x^2 - x - 2) &< 0 \\ (x - 2)(x + 1) &< 0 \end{aligned} \]

To find the interval where the expression is negative, we examine the sign of the factors around the roots.

Figure: A number line showing the interval between -1 and 2 where the quadratic inequality is satisfied.

So the interval is:

\[ -1 < x < 2 \]

Singularities and Tangents

At \( x = -1 \) and \( x = 2 \), the derivative \[ y' = \frac{1-2x}{y} \] is undefined.

Vertical Tangent

on \( y(x) \)

Stay away from there

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Direction Field and Solution Curve

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Differential Equation

\[ y' = \frac{1-2x}{y} \]

The direction field plotter shows the slope of the solution at various points in the \(xy\)-plane. A specific solution curve is highlighted, passing through an initial condition (IC).

Figure: A direction field plot with a red solution curve, an initial condition point labeled IC, and a blue arrow labeled Solution.

Observations:

The red curve represents the particular Solution.
The point marked IC indicates the Initial Condition.
The slopes become vertical as the curve approaches the x-axis where \( y = 0 \).

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Homogeneous equations

If \( \frac{dy}{dx} \) can be expressed as a function of \( \frac{y}{x} \)

example:

\[ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \]\[ = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 \]

Homogeneous eqs can be turned into separable by making a change of variable \( y = x v(x) \) or \( v(x) = \frac{y}{x} \)

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\[ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 \]

let \( y = x v(x) \) so \( \frac{dy}{dx} = \frac{d}{dx}(xv) \) product rule

\[ = x \frac{dv}{dx} + v \]

left side of DE

right side is \( 1 + v + v^2 \)

new DE in v:

\[ x \frac{dv}{dx} + v = 1 + v + v^2 \]\[ x \frac{dv}{dx} = 1 + v^2 \]

separable!

\[ \frac{1}{1+v^2} dv = \frac{1}{x} dx \]

integrate

\[ \tan^{-1}(v) = \ln|x| + C \]

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\[ v = \tan(\ln|x| + c) \quad y = xv \]

so

\[ y = x \tan(\ln|x| + c) \]

homogeneous DE:

\[ \frac{dy}{dx} = f\left(\frac{y}{x}\right) \]

so slope of \( y \) only depends on \( \frac{y}{x} \)

for example, slope at \( (2, 2) \) and \( (4, 4) \) are the same

The diagram illustrates that points along the same line through the origin share the same slope in a homogeneous differential equation.

Figure: Coordinate axes with a line through the origin; red marks show identical slopes at different points.

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Direction Field Visualization

Figure: A direction field plotter grid with arrows representing slopes across a Cartesian plane.

any line thru origin will have same slope