Category: First-Order ODE

Separable Differential Equations

Consider the differential equation `y'=f(t,y)`, or `(dy)/(dt)=f(t,y)`.

If the function `f(t,y)` can be written as the product of the function `g(t)` (function that depends only on `t`) and the function `u(y)` (function that depends only on `y`), such a differential equation is called separable.

Homogeneous Equations

If in the differential equation `y'=f(t,y)`, the function `f(t,y)` has the property that `f(at,ay)=f(t,y)`, such a differential equation is called homogeneous.

It can be transformed into a separable equation using the substitution ` y=ut` along with the corresponding derivative `(dy)/(dt)=(du)/(dt)t+u`. The resulting equation is solved as a separable equation, and the required solution is obtained by back subtitution.

Exact Equations

The differential equation `M(x,y)dx+N(x,y)dy=0` is exact, if there exists a function `f` such that `df=M(x,y)dx+N(x,y)dy`.

In this case, the equation can be rewritten as `df=0`, which gives the solution `f=C`.

Linear Differential Equations

A first-order linear differential equation has the form `y'+p(t)y=q(t)`.

To solve it, rewrite it in the differential form: `(dy)/(dt)+p(t)y=q(t)`, or `(p(t)y-q(t))dt+dy=0`.

Now, using the facts about exact equations, we see that this equation is not exact. However, since `M(t,y)=p(t)y-q(t)` and `N(t,y)=1`, we have that `(partial M)/(partial y)=p(t)` and `(partial N)/(partial x)=0`. So, `1/N((partial M)/(partial y)-(partial N)/(partial x))=1/1 (p(t)-0)=p(t)`. Thus, the integrating factor is `I(t,y)=e^(int p(t) dt)`.

Bernoulli Equations

A Bernoulli equation has the form `y'+p(t)y=q(t)y^n` where `n` is a real number.

Using the substituion `z=y^(1-n)`, this equation can be transformed into a linear one.

Example 1. Solve `y'-3/ty=t^4y^(1/3)`.