An equation of the form t2 d2y dt2 + αt dy dt + βy = 0, t > 0, (1) where α and β are real constants, is called an Euler equation. If we let x = ln t and calculate dy/dt and d2y/dt2 in terms of dy/dx and d2y/dx2, then equation (1) becomes d2y dx2 + (α - 1) dy dx + βy = 0. (2) Observe that equation (2) has constant coefficients. If y1 (x) and y2 (x) form a fundamental set of solutions of equation (2), then y1 (ln t) and y2 (ln t) form a fundamental set of solutions of equation (1). Use the method above to solve the given equation for t > 0. t2y'' + 5ty' + 3y = 0

The solution to the Euler equation

t²y'' + 5ty' + 3y = 0

is

y = C1/t + C2/t³

Step-by-step explanation:

We are required to solve the Euler equation

t²y'' + 5ty' + 3y = 0 ... (1)

Let x = lnt ... (2)

=> t = e^x

Let d/dx be D

dy/dt = dy/dx * dx/dt

= (1/t) dy/dx

=> tdy/dt = dy/dx = Dy ... (3)

Again, d²y/dt² = d/dt (dy/dt) = d/dt[ (1/t) dy/dx] = (-1/t²) dy/dx + (1/t) d²y/dx² * dx/dt

= (-1/t²) (d²y/dx² - dy/dx) = (1/t²) (D² - D)

t²d²y/dt² = D (D - 1) y ... (4)

Using (3) and (4) in (1)

t²y'' + 5ty' + 3y = D (D - 1) y + 5Dy + 3y = 0

(D² - D + 5D + 3) y = 0

(D² + 4D + 3) y = 0

The auxiliary equation is

m² + 4m + 3 = 0

(m + 1) (m + 3) = 0

m = - 1, - 3

y = C1e^ (-x) + C2e^ (-3x)

But t = e^x

So,

y = C1/t + C2/t³