How to Solve a Linear Second Order Differential Equation

Differential equations have many applications to physics and engineering. A second order linear differential equation has the form

y''(t) + ay'(t) + by(t) = 0,

where a and b are constants. If the equation has initial conditions y(0) = c and y'(0) = d, then the equation is called an "initial value problem" and it can be solved completely. That is, you can find an explicit equation for y(t).

The form of the equation depends on solution to the associated quadratic equation

z² + az + b = 0

and there are three possible cases. You can use the method outlined below to solve these kinds of equations, or use the differential equation solver on the left.

Case I: a² - 4b > 0

When a² - 4b > 0, the equation has the form

y(t) = k₁e^pt + k₂e^qt,

where p = -0.5a + 0.5sqrt(a² - 4b) and q = -0.5a - 0.5sqrt(a² - 4b). The constants k₁ and k₂ depend on the initial conditions of the equation. When you plug in the initial conditions into y(t) and y'(t), it creates a system of two linear equations in the variables k₁ and k₂. Thus you can find the complete solution.

If one of p or q is zero, the solution equation simplifies to y(t) = k₁e^pt + k₂.

Case II: a² - 4b = 0

When a² - 4b = 0, the equation has the form

y(t) = k₁e^pt + k₂te^pt,

where p = -0.5a. As before, the values of the constants k₁ and k₂ can be determined from the initial conditions.

Case III: a² - 4b < 0

When a² - 4b < 0, the equation has the form

y(t) = e^pt(k₁sin(rt) + k₂cos(rt)),

where p = -0.5a and r = 0.5sqrt(4b - a²).

Degenerate Case

If both a and b equal zero, the differential equation solution has the form

y(t) = k₁t + k₂.

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