How to Solve a Second Order Euler Differential Equation

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

t²y''(t) + aty'(t) + by(t) = 0,

where a and b are constants, is called an Euler equation. If the equation has initial conditions y(1) = c and y'(1) = d, then the initial value problem 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 + (a-1)z + b = 0

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

Case I: (a-1)² - 4b > 0

When (a-1)² - 4b > 0, the equation has the form

y(t) = k₁t ^p + k₂t ^q,

where p = 0.5(1-a) + 0.5sqrt((a-1)² - 4b) and q = 0.5(1-a) - 0.5sqrt((a-1)² - 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) = k₁t ^p + k₂.

Case II: (a-1)² - 4b = 0

When (a-1)² - 4b = 0, the equation has the form

y(t) = k₁t ^p + k₂t ^pLn|t|,

where p = 0.5(1-a). As before, the values of the constants k₁ and k₂ can be determined from the initial conditions of the problem.

If a = 1 and b = 0, the differential equation solution has the form

y(t) = k₁ + k₂Ln|t|.

Case III: (a-1)² - 4b < 0

When (a-1)² - 4b < 0, the solution to the Euler equation has the form

y(t) = t ^p(k₁sin(rLn|t|) + k₂cos(rLn|t|)),

where p = 0.5(1-a) and r = 0.5sqrt(4b - (a-1)²).

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