Techniques

I’ve found this video series to be very helpful

Formula for Linear Separable First order ODE

1. Write in standard form

$$y^{\prime }+p(x)y=q(x)$$

2. Determine integrating factor

$$I(x)=e^{\int p(x)dx}$$

3. General solution

$$y(x)=\frac{1}{I(x)}\left(\int I(x)Q(x)dx+C\right)$$

Only add the integrating factor here, at the end

Bernoulli Equations

In the form:

$$y^{\prime }+p(x)y=f(x)y^r$$

Where r is any real number other than 0 or 1.

1. Divide by $y^r$

$$y^{\prime }{y}^{-r}+p(x)y^{1-r}=f(x)$$

2. Substitute $v=y^{1-r}$ and it’s derivative $\frac{dv}{dx}=(1-r)y^{-r}\frac{dy}{dx}$ Following U-Sub rules.. $\to \frac{dy}{dx}=\frac{dv}{dx}(1-r)y^r$

$$\frac{dv}{dx}(1-r)y^r{y}^{-r}+p(x)v=f(x)$$

$y^r$ and $y^{-r}$ cancel..

$$\frac{dv}{dx}(1-r)+p(x)v=f(x)$$

3. This is now linear, first order, and separable. Solve as such
4. Back substitute, and solve for y

Higher Order ODE’s

Existence & Uniqueness

For $y^{\prime \prime }+p(x)y^{\prime }+q(z)y=f(x)$ And $y(a)=b_0,y^{\prime }(x)=b_1$

If $p,q,f$ are continuous on an interval I containing a, then there exists a solution on I which is unique

Superposition

For $y^{\prime \prime }+p(x)y^{\prime }+q(z)y=0$ There are necessarily two general solutions, and any linear combination of them is also a solution. Additionally, if $y_1$ and $y_2$ are linearly independent, then all solutions are a linear combination of them.

Homogeneous With Constant Coefficients

For $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ Substitute $y=e^{rt}\to a{r}^2+br+c=0$ Solve as a quadratic, by factoring or by using the quadratic formula.

You will at this point have either two real and distinct values for r, or a repeated real root, or complex roots.

Distinct Real

$y=C_1{e}^{r_{1}t}+C_2{e}^{r_{2}t},y^{\prime }=C_1{r}_1{e}^{r_{1}t}+C_2{r}_2{e}^{r_{2}t}$ Then just plug in your two initial values (one for $y$, one for $y^{\prime }$), and solve for $C_1$, $C_2$

Repeated Real

This occurs when $b^2-4ac=0$ $y=C_1{e}^{rt}+C_{2}t{e}^{rt},y^{\prime }=C_{1}r{e}^{rt}+C_2(1+rt)e^{rt}$

Complex Roots

This occurs when $b^2-4ac<0$ the roots are: $r=\alpha \pm i\beta$ So the two general solutions are: $y_1=e^{\alpha t}(\cos \beta t+i\sin \beta t)$ $y_2=e^{\alpha t}(\cos \beta t-i\sin \beta t)$ $\frac{y_1+y_2}{2i}$ is a solution because it is a linear combination of solutions. This leads to: $e^{\alpha t}\sin \beta t$ and $e^{\alpha t}\cos \beta t$ as valid general solutions, which are linearly independent, and real valued. So a specific solution takes the form: $y=C_1{e}^{\alpha t}\cos (\beta t)+C_2{e}^{\alpha t}\sin (\beta t)$ and $y^{\prime }=(\alpha C_1+\beta C_2)e^{\alpha t}\cos (\beta t)+(\alpha C_2-\beta C_1)e^{\alpha t}\sin (\beta t)$

Non Homogeneous with constant coefficients (Undetermined Coefficients)

The sum of the solution to the homogeneous variation of the ODE and any particular solution is a solution to the non-homogeneous ODE

1. Find the homogeneous solution
2. Remove the coefficient from the right side, then plug it in with an undetermined coefficient in place, to find the value of that efficient. For: $y^{\prime \prime }-2y^{\prime }-3y=3e^{2t}$ Plug in $A{e}^{2t}$: $(A{e}^{2t}{)}^{\prime \prime }-2(A{e}^{2t}{)}^{\prime }-3A{e}^{2t}=3e^{2t}$ Derrive: $4A{e}^{2t}-4A{e}^{2t}-3A{e}^{2t}=3e^{2t}$ $A=-1$ So one particular solution is: $y_p=-e^{2t}$

Since the solution to the homogeneous is: $y_h=c_1{e}^{3t}+c_2{e}^{-t}$, the general solution is: $y_g=c_1{e}^{3t}+c_2{e}^{-t}-e^{2t}$

Remember that you must solve for the general solution before implementing the initial value.

Also, if the right side matches the form of any term of the homogeneous solution, then try $At$ as the undetermined coefficient, instead of just $A$

Variation of Parameters (for second order)

Used when the coefficients of $y$, $y^{\prime }$ etc are not constants.

1. Solve the homogeneous
2. Guess that $y_g=u_1{y}_1+u_2{y}_2$
3. Use the formulas: Where $g$ is a function of x (the right side of the equation), when the coefficient of $y^{\prime \prime }$ is 1 $u_1=\int \frac{y_{2}g}{y_1{y}_2^{\prime }-y_2{y}_1^{\prime }}$ $u_2=\int \frac{y_{1}g}{y_1{y}_2^{\prime }-y_2{y}_1^{\prime }}$

The Wronskian

A way to figure if functions are linearly independent Form a square matrix and take it’s determinate: $\det \left[\begin{array}{ccc}{y}_1& y_2& y_3\\ y_1^{\prime }& y_2^{\prime }& y_3^{\prime }\\ y_1^{\prime \prime }& y_2^{\prime \prime }& y_3^{\prime \prime }\end{array}\right]$ If it’s determinate is non-zero in at least one place, then they are linearly independent.