Solve the given differential equation. x²y'' + xy' + 4y = 0

Solution:

Since the given equation is a second-order differential equation we can write the auxiliary equation for it as:

m²x² + mx + 4 = 0

The roots of the equation are

= -m ± √m² - 4(m²)⁴ / 2m²

= -m ± √m² - 16m² / 2m²

= -m ± √-15m² / 2m²

= -m ± im√15 / 2m²

-1/2m + i√15/2m and -1/2m - i√15/2m

The general solution to the differential equation is: \((c_{1}cosx + c_{2}Sinx)e^{\dfrac{\sqrt{15}x}{2m}}\)

Solve the given differential equation. x²y'' + xy' + 4y = 0

Summary:

Solving the given differential equation. x²y'' + xy' + 4y = 0 for a general equation we get \((c_{1}cosx + c_{2}Sinx)e^{\sqrt{15}x/2m}\)