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2.2 Separable Diff. Eqs.

let's revisit \(\frac{dy}{dx} = xy\)

we solved by using calculus

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

let \(u = y\) then \(du = \frac{dy}{dx} dx\)

integrate both sides

\[\int \frac{1}{y} \frac{dy}{dx} dx = \int x dx\]
\[\int \frac{1}{u} du = \int x dx\]
\[\ln |u| = \frac{1}{2}x^2 + C\]
\(\leftrightarrow\)
\[\ln |y| = \frac{1}{2}x^2 + C\]
\[|u| = e^C e^{\frac{1}{2}x^2}\]
\[u = \pm e^C e^{\frac{1}{2}x^2}\]
\[y = Ce^{\frac{1}{2}x^2}\]
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notice it almost looked like we did this:

\[\frac{dy}{dx} = xy\]

"multiply by \(dx\)" and "divide by \(y\)"

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

integrate

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

actually NOT mathematically "legal"

\(\frac{dy}{dx}\) is a notation for \(y'\)

it is NOT a quotient

BUT, we get the right result, so for the purpose of solving diff. eqs. we will pretend it's "legal"

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Separable Differential Equations

\[ \frac{dy}{dx} = f(x,y) \] is separable if we can separate \( f(x,y) \) into a product or quotient of a function of \( x \) and a function of \( y \).

All separable diff. eqs. can be solved the same way as the previous example.

Examples of Separable:

\[ \frac{dy}{dx} = \frac{\sin(x)}{y} = \frac{g(x)}{h(y)} \]
\[ \frac{dy}{dx} = e^y \ln|x| = g(y)h(x) \]
\[ \frac{dy}{dx} = y = y \cdot 1 \]
separable
also linear
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Let's solve

\[ \frac{dy}{dx} = \frac{\sin(x)}{y} \]

Separate \( x \) and \( y \)

\[ y \, dy = \sin(x) \, dx \]

Integrate

\[ \int y \, dy = \int \sin(x) \, dx \]
\[ \frac{1}{2} y^2 = -\cos(x) + C \]

Implicit form of solution

\( \cos(x) + \frac{1}{2} y^2 = C \)

or

\[ 2 \cos(x) + y^2 = C \]

Explicit form

\[ y^2 = -2 \cos(x) + C \]
\[ y = \pm \sqrt{C - 2 \cos(x)} \]

resolved by initial condition \( y(x_0) = y_0 \)

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for example, let's say \( y(0) = 3 \)

\[ y = \pm \sqrt{C - 2 \cos(x)} \]\[ 3 = \pm \sqrt{C - 2 \cos(0)} \]\[ 3 = \pm \sqrt{C - 2} \]

\( C = 11 \) and we must choose \( + \)

particular solution : \( y = \sqrt{11 - 2 \cos(x)} \)

\[ \frac{dy}{dx} = \frac{3x^2 - e^x}{2y - 5} \quad y(0) = 1 \]

ALWAYS separate by multiplication or division. NEVER by addition/subtraction

\[ (2y - 5) dy = (3x^2 - e^x) dx \]\[ \int (2y - 5) dy = \int (3x^2 - e^x) dx \]
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\[ y^2 - 5y = x^3 - e^x + C \]\[ y(0) = 1 \]\[ 1 - 5 = 0 - 1 + C \quad C = -3 \]

can find C using initial condition any time AFTER integration

\[ y^2 - 5y = x^3 - e^x - 3 \]

implicit form

NOT all diff. eqs. have solutions in explicit form.

here, we can complete the square on the left

\[ y^2 - 5y + \left( \frac{-5}{2} \right)^2 = x^3 - e^x - 3 + \left( \frac{-5}{2} \right)^2 \]\[ y^2 - 5y + \frac{25}{4} = x^3 - e^x + \frac{13}{4} \]\[ \left( y - \frac{5}{2} \right)^2 = x^3 - e^x + \frac{13}{4} \]
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\[ y - \frac{5}{2} = \pm \sqrt{x^3 - e^x + \frac{13}{4}} \]
\[ y = \frac{5}{2} \pm \sqrt{x^3 - e^x + \frac{13}{4}} \]

resolve this

\[ y(0) = 1 \]
\[ 1 = \frac{5}{2} \pm \sqrt{\frac{9}{4}} = \frac{5}{2} \pm \frac{3}{2} \]
\[ y = \frac{5}{2} - \sqrt{x^3 - e^x + \frac{13}{4}} \]

On what interval of \( x \) is the solution valid?

one way: find domain of \( y \)

\[ \text{here, } x^3 - e^x + \frac{13}{4} \geq 0 \]

\( x = ? \)

not easy equation to solve

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Alternative: look at where the slope of \( y \) is vertical

these bound intervals on which \( y \) is valid

DON'T differentiate \( y \)!

use the diff. eq. \( \frac{dy}{dx} \) to construct a slope field

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Slope Field and Solution Curves

\[ \frac{dy}{dx} = \frac{\sin(x)}{y} \]
Slope field for dy/dx = sin(x)/y showing vector directions and red solution curves, including one passing through (0,1).