1 10 points Consider the differential equation t2y 3ty y

1. (10 points) Consider the differential equation t^2y + 3ty? + y = 0, t > O. (c) Write the general solution of the DE.

Solution

t²y\" + 3ty\' + y = 0, t>0 is Euler\'s equation.

Let x = lnt

Then dy/dt = (dy/dx)(dx/dt)...Using chain rule of derivatives

=> dy/dt = (dy/dx)(1/t) (because dx/dt = 1/t)

Now d²y/dt² = (d²y/dx²)(dx/dt)²+(dy/dx)(d²x/dt²) = (1/t²)(d²y/dx²)+(-1/t²)(dy/dx) = (1/t²)(d²y/dx²-dy/dx)

Substitute these values in the original equation, we get:

t²(1/t²)(d²y/dx²-dy/dx)+3t(1/t)(dy/dx)+y = 0

=> d²y/dx²-dy/dx+3dy/dx+y = 0

=> d²y/dx² + 2dy/dx + y = 0

This is ordinary differential equation whose characteristic equation is given by:

r² + 2r + 1 = 0 => (r+1)² = 0 => r=-1, -1

So the general solution of this ordinary differential equation in x is given by:

y = Ae^-x + Bxe^-x

Now substitute x=lnt

y = Ae^-lnt + Blnt e^-lnt = Ae^{lnt^-1} + Blnte^lnt^-1

=> y = A(1/t)+Blnt(1/t) = (1/t)(A+Blnt)

=> y = (1/t)(A+Blnt)