Week 2 - Integrals as Solutions

Solving Differential Equations

When solving a differential equation, e.g.

\[\frac{dy}{dx}=5y\]

our goal is to find a function $y(x)$ that satisfies the differential equation.

Verifying Differential Equations

It is relatively straightforward to verify that a function is a solution to a differential equation. For example:

Verify that $x(t)=sin(t)+cos(t)$ is a solution to the differential equation $\frac{dx}{dt}+x=2cos(t)$

1. Take the derivative of the proposed solution: \(\frac{d}{dt}(sin(t)+cos(t)) + (sin(t)+cos(t))\\ = cos(t) - sin(t) + sin(t) + cos(t)\\ = 2cos(t)\)
2. Thus, $x(t)$ is indeed a solution to the differential equation $\frac{dx}{dt}+x=2cos(t)$ because it satisfies the differential equation.

## Solving First Order Differential Equations Some first order differential equations take the form $\frac{dy}{dx}=f(x)$ or $\frac{dy}{dx}=g(y)$. This type of differential equation is trivial to solve, by integrating both sides of the equation. For example:

Solve the differential equation $y’=4x^3$

1. Integrate both sides: $\int y’ dx = \int 4x^3 dx$
2. Therefore $y(x) = x^4+C$

The equation $y(x)=x^4+C$ is the general solution to this differential equation.

Initial Conditions

Sometimes, an initial condition might be specified. This allows us to solve for $C$ and thereby arrive at a particular solution. For example:

Solve the differential equation $y’=4x^3$ such that $y(0)=1$

1. Integrate both sides: $\int y’ dx = \int 4x^3 dx$
2. Therefore $y(x) = x^4+C$
3. Evaluate the general solution at the condition: $y(0) = 1 = (0)^4 + C$
4. Solve for C: $C=1$
5. Thus the particular solution is $y(x)=x^4+1$

Trivial Differential Equations

Consider equations of the form \(\frac{dy}{dx}=f(y)\) where $f(y)=0$. In this case, the solution is trivial: $y(x)=C$.

If, however, $f(y)\neq 0$ then \(\int \frac{dy}{f(y)}=\int dx + C\\ = \int \frac{dy}{f(y)} = x + C\)

For example:

Consider $y’=5y$ such that $y(0)=2$. Find the particular solution.

1. Rewrite: $\frac{dy}{dt}=5y$
2. Rearrange: $\frac{dy}{y}=5 dt$
3. Integrate: $\int \frac{dy}{y} = \int 5 dt$
4. Solve: \(\ln|y|=5t+C\\ \textrm{Exponentiate: }e^{ln|y|}=e^{5t+C}\\ |y|=e^{5t}\cdot e^C\\ \textrm{Observe } e^C \textrm{ is constant}\\ |y|=Ce^{5t}\\ \therefore y(t) = \pm Ce^{5t}\)
5. Using the initial condition, $y(0)=2=C\cdot 1 \implies C=2$

Singular Solutions

Differential equations may have singular solutions: constant-value solutions which don’t satisfy the general solution. E.g., $y \equiv 0$ is a solution to the differential equation $\frac{dy}{dx}=2xy^2$ whose general solution is $\frac{1}{C-x^2}$; observe $\frac{1}{C-x^2}\neq 0$ for all $C \in \real$.