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 \(y^{''}+y^{'}-2y=0\).
First, we can determine the first and second derivatives of \(y(t)\):
\(y^{'} = -2e^{-2t}\)
\(y^{''} = 4e^{-2t}\)
Now plug into the DE:
\(y^{''}+y^{'}-2y=0\)
\(4e^{-2t}-2e^{-2t}-2e^{-2t}=0\)
\(0=0\)
Therefore, we can determine 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. | \(y=\cfrac{1}{6} e^{3t}\) |
| 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 \((y^{'})^2 + 1 = -yy^{''}\).
First, we can implicity find the second derivative with respect to \(x\):
\(\cfrac{d^2}{dx^2} (x^2 + y^2 ) = \cfrac{d^2}{dx^2} (1) \)
\(\cfrac{d}{dx} (2x + 2yy^{'} ) = \cfrac{d}{dx} (0) \)
\(2 + 2yy^{''} + 2y^{'}y^{'} = 0 \)
\((y^{'})^2 = -1 -yy^{''} \)
Next, we can plug into the DE to confirm it is a solution:
\((y^{'})^2 + 1 = -yy^{''}\)
\((-1-yy^{''}) + 1=-yy^{''}\)
\(-yy^{''}=-yy^{''}\)
Therefore, we can determine that \(x^2 + y^2 = 1\) is a solution to \((y^{'})^2 + 1 = -yy^{''}\).
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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:
\(y(x_0) = y_0, \cfrac{dy}{dx}(x_0) = y_1, \cfrac{d^2y}{dx^2}(x_0) = y_2\)
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:
\(c_1 = \cfrac{3}{5}, c_2 = \cfrac{2}{5}\)
\(y = \cfrac{3}{5} e^{2x} + \cfrac{2}{5} e^{-3x}\)