Thus far we have looked at first-order differential equations. A higher order differential equation will have second, third and up to n derivatives:

\( \cfrac{d^3y}{dx^3} - x\cfrac{d^2y}{dx^2} + 2 = \sin {x} \)

The general form of a higher-order DE is:

\( a_n(x)\cfrac{d^ny}{dx^n} + a_{n-1}(x)\cfrac{d^{n-1}y}{dx^{n-1}} + \dots + a_1(x)\cfrac{dy}{dx} + a_0(x)y = g(x)\)

Notice that the coefficients infront of the derivatives are functions of \(x\). For an intial value problem, the DE is subject to:

\( y(x_0) = y_0, \cfrac{dy}{dx}(x_0) = y_1, \dots \cfrac{d^{n-1}y}{dx^{n-1}} (x_0) = y_{n-1}\)

When \(g(x) = 0\), the equation is **homogeneous**, otherwise the equation is **non-homogeneous**.

Classify the DE \( \ddot{y} + 2x\dot{y} - e^{x^2} = 0 \).

Show Answer
The equation has an **order** of 2 since the highest derivative is the second derivative of \(y\). It is **non-homogeneous** because
\(g(x) = e^{x^2}\).

## Superposition Principle

the **Superposition Principle** states that if \(y_1, y_2, \dots y_n\) are solutions to a homogeneous differential equation on interval \(I\), then so is
any linear combination of those solutions:

\( y = \sum {c_iy_i} \)

Show that a linear combination of solutions \(y = e^{-10x}\) and \(y = e^{5x}\) are also solutions to the DE \(y^{''} + 5y^{'} - 50 y = 0 \).

Show Answer
First we can show these are indeed solutions to the DE:

\( (100e^{-10x}) + 5 (-10e^{-10x}) - 50 (e^{-10x}) = 0 \)

\(100e^{10x} - 50e^{-10x} - 50 e^{-10x} = 0 \)

\(0 = 0 \)

\( (25e^{5x}) + 5 (5e^{5x}) - 50 (e^{5x}) = 0 \)

\(25e^{5x} + 25e^{5x} - 50 e^{5x} = 0 \)

\(0 = 0 \)

Now let's try a linear combination of the two with arbitrary constant coefficients \( y= c_1 e^{-10x} + c_2 e^{5x} \). Take derivatives and plug into the DE:

\( (100c_1 e^{-10x} + 25c_2 e^{5x}) + 5 (-10c_1 e^{-10x} + 5c_2 e^{5x}) - 50 (c_1 e^{-10x} + c_2 e^{5x}) = 0 \)

\( 100c_1 e^{-10x} + 25c_2 e^{5x} - 50c_1 e^{-10x} + 25c_2 e^{5x} - 50 c_1 e^{-10x} - 50 c_2 e^{5x} = 0 \)

\( c_1 (100e^{-10x} - 50 e^{-10x} - 50e^{-10x} ) + c_2 (25 e^{5x} + 25 e^{5x} - 50 e^{5x} ) \)

\( c_1 (0 ) + c_2 (0 ) \)

\(0 = 0 \)