Solutions to Differential Equations

A solution to a differential equation is a function that satisfies the DE. The solution must be continueously differentiable on an interval, I. You can check the solution by taking the appropriate derivatives and confirming the LHS = RHS.

Confirm that \(y(t)=e^{-2t}\) is a solution to \(y^{''}+y^{'}-2y=0\).

---

There are different type of solutions discussed below:

Type	Description	Example
Trival Solution	The solution \(y=0\) is a trivial solution.	\(y=0\)
General/Family Solution	Solution with constants that can be solved given the initial value. Sometimes refered to as a familiy of solution as any constant will solve the DE.	\(y=c_1 e^{3t}\)
Particular Solution	Solution to a particular set of initial values.	\(y=\frac{1}{6} e^{3t}\)
Implicit Solution	Solution that is not defined explicitly as \(y = y(x)\).	\(\sin x + \cos y = 1\)

Show that \( x^2 + y^2 = 1\) is a solution to the DE \( (y^{'})^2 + 1 = -yy^{''}\).

---

Initial Value Problems

To solve for the particular solution to a DE, often an initial value is provided to allow us to solve for the constants. For a DE of order n, initial values must be provided up to the n-1 derivative. For exmaple, for DE such as \(\cfrac{d^3y}{dx^3}=f(x,y,dy / dx, d^2y / dx^2) \) must be subject to initial conditions \(y(x_0) = y_0, dy / dx (x_0) = y_1, d^2y / dx^2 (x_0) = y_2\).

Consider the general solution \(y = c_1 e^{2x} + c_2 e^{-3x}\) to the DE \( y^{''}+y^{'}-6y=0 \). We can show that this is indeed a solution:

\( y^{''}+y^{'}-6y=0 \)

\( (4c_1 e^{2x} + 9c_2 e^{-3x})+ (2c_1 e^{2x} -3c_2 e^{-3x}) - 6(c_1 e^{2x} + c_2 e^{-3x})= 0 \)

\( 0 = 0 \)

This is true for any value of \( c_1, c_2\). Shown graphically below:

Now we can find the particular solution given the contraints \( y(0) = 1, y^{'} (0) = 0\)

\(y = c_1 e^{2x} + c_2 e^{-3x}\)

\(y(0) = c_1 e^{2(0)} + c_2 e^{-3(0)}\)

\(1 = c_1 + c_2 \)

\( y{'} = 2c_1 e^{2x} -3c_2 e^{-3x}\)

\( y{'}(0) = 2c_1 e^{2(0)} -3c_2 e^{-3(0)}\)

\(0 = 2c_1 - 3c_2 \)

Solving the two equations for two unknowns yields:

\( c_1 = \frac{3}{5}, c_2 = \frac{2}{5}\)

\( y = \frac{3}{5} e^{2x} + \frac{2}{5} e^{-3x}\)