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 \( \ddot{y} + 2x\dot{y} - e^{x^2} = 0 \).
The equation has an order of 2 since the highest derivative is the second derivative of \(y\). It is non-homogeneous because \(g(x) = e^{x^2}\).
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 \(y^{''} + 5y^{'} - 50 y = 0 \).
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:
\( (100c_1 e^{-10x} + 25c_2 e^{5x}) + 5 (-10c_1 e^{-10x} + 5c_2 e^{5x}) - 50 (c_1 e^{-10x} + c_2 e^{5x}) = 0 \)
\( 100c_1 e^{-10x} + 25c_2 e^{5x} - 50c_1 e^{-10x} + 25c_2 e^{5x} - 50 c_1 e^{-10x} - 50 c_2 e^{5x} = 0 \)
\( c_1 (100e^{-10x} - 50 e^{-10x} - 50e^{-10x} ) + c_2 (25 e^{5x} + 25 e^{5x} - 50 e^{5x} ) \)
\( c_1 (0 ) + c_2 (0 ) \)
\(0 = 0 \)