4 L4: ODE Separation of Variables (2.1-2.2 part `a`.)

4.1 ODE Separation of variables (case I)

\[\boxed{\frac{dy}{dx}=f(x)}\]

\[\begin{align} dy&=f(x)\,dx\quad(\text{differential form})\\ \int dy &= \int f(x)\,dx\\ y&=\int f(x)\,dx \end{align}\]

In principle we can integrate the RHS.

Q: Is every function integrable?

A: No. (A lot are though)

4.2 Example

Solve: \[\frac{dy}{dx}=\cos x\]

Solution

Sol: \[\begin{align} \int dy &=\int \cos x\, dx\\ y&= \sin x +C \end{align}\]

4.3 ODE Separation of variables (case II)

\[\boxed{\frac{dy}{dx}=f(x,y)}\] We would like to have $f(x,y)=g(x)h(y)$,

if true, $\Rightarrow$ \[\begin{align} \frac{dy}{dx}&=g(x)h(y)\\ \frac{1}{h(y)}dy&=g(x)\,dx\quad(\text{Separate variable})\\ \int \frac{1}{h(y)}dy&=\int g(x)\,dx \end{align}\]

Warning:

Solution may not be valid for all $(x,y)$ values due to singularities.

Q: Can every function $f(x,y)$ be separated into $g(x)h(y)$ ?

A: No.

4.4 Definition: Separable

Def:

If the $1^{st}$ order diff. eq$^n$ can be written in the form $f(y)\,dy=g(x)\,dx$ it is said to be separable

4.5 Example

Solve: \[\boxed{\frac{dy}{dx}=xy}\]

Solution

\[\begin{align} \frac{1}{y}\,dy&=x\,dx,\qquad(y\neq 0)\\ \int\frac{1}{y}\,dy&=\int x\,dx\\ \ln|y|&=\frac{x^2}{2}+C\\ |y|&=e^Ce^{x^2/2},\qquad e^C>0\\ |y|&=De^{x^2/2},\qquad D>0 \end{align}\]

Note: Additional Solution $y=0$

Q: Are there any other solutions ?

A: Maybe (Next time).

4.6 More examples

4.6.1 Example 1:

Solve: \[\frac{dy}{dx}=\frac{2xy}{1+x^2}\]

Solution

\[\begin{align} \frac{dy}{dx}&=\frac{2xy}{1+x^2}\\ \frac{1}{y}\,dy&=\frac{2x}{1+x^2}\,dx\\ \int \frac{1}{y}\,dy&=\int \frac{2x}{1+x^2}\,dx\\ \text{(left side trivial,}&\text{right side simple $u$ substitution)}\\ \ln |y| + C_1 &= \int \frac{1}{u}\,du\\ \ln |y| + C_1 &= \ln |u| +C_2\\ \ln |y| &= \ln |1+x^2| +C\\ \text{Let's recall }& C=\ln C_3\\ \Rightarrow\\ \ln |y| &= \ln(1+x^2) +\ln C_3\\ \text{(we know:)}& \qquad(\log (a\cdot b)=\log a+\log b)\\ \Rightarrow\\ \ln |y| &= \ln\left((1+x^2)\cdot \ln C_3\right)\\ |y|&=(1+x^2)\cdot C \end{align}\]

4.6.2 Example 2: Detailed step by step

Solve:

\[\frac{dy}{dx} + 18xy = 0\]

Solution

Separate Variables: \[\frac{dy}{y} = -18x \, dx\]
Integrate Both Sides: \[\int \frac{1}{y} \, dy = \int -18x \, dx\]

\[\ln|y| = -9x^2 + C\]

(where $C$ is the constant of integration)
Solve for $y$: Taking the exponential of both sides: \[|y| = e^{-9x^2 + C}\]

Since $|y|$ can be positive or negative, we have two cases:
- Case 1: $y = e^{-9x^2 + C}$
- Case 2: $y = -e^{-9x^2 + C}$
Simplifying the constant of integration: \[y = \pm e^{-9x^2 + C/2}\]

(where $C/2$ is another constant)

Therefore, the general solution to the differential equation is:

\[y = A e^{-9x^2} \quad \text{or} \quad y = -A e^{-9x^2}\]

4.6.3 Example 3: Implicit Solutions

Find implicit solutions of \[y'=\frac{2x + 1}{5y^4 + 1},\qquad y(2) = 1\]

Solution

\[\begin{align} \frac{dy}{dx}&=\frac{2x + 1}{5y^4 + 1}\\ (5y^4 + 1)\,dy&=(2x+1)\,dx\\ \int (5y^4 + 1)\,dy&=\int (2x+1)\,dx\\ y^5+y+C_1&=x^2+x+C_2,\qquad(\text{absorving})\\ y^5+y&=x^2+x+C \end{align}\]

Now: $y(2)=1$, then:

\[\begin{align} y^5+y&=x^2+x+C\\ 1^5+1=2^2+2+C\\ \therefore C&= -4\\ \end{align}\]

Finally the implicit solution for the diffyQ with i.c. is:

\[\boxed{y^5+y=x^2+x-4}\]