How do you solve the initial value with 4y4yy0 y01 y015Solut

How do you solve the initial value with 4y\'\'-4y\'+y=0, y(0)=1, y\'(0)=-1.5?

4y\'\'-4y\'+y=0 by characteristic eqn, 4m^2-4m+1 = 0 ==> (2m-1)^2 = 0 m1 = 1/2, m2 = 1/2 y = c1e^(t/2) + c2te^(t/2) at y(0) = 1 1 = C1 + 0C2 ==> C1 = 1 y\' = (1/2)c1e^(t/2) + c2e^(t/2) + (t/2)c2e^(t/2) at y\'(0) = -1.5, -1.5 = (1/2)c1 + c2 ==> -1.5 = (1/2)(1) + c2 ==> c2 = -5/2 y = e^(t/2) -(5/2) te^(t/2)

$How do you solve the initial value with 4y\'\'-4y\'+y=0, y(0)=1, y\'(0)=-1.5?Solution 4y\'\'-4y\'+y=0 by characteristic eqn, 4m^2-4m+1 = 0 ==> (2m-1)^2 = 0 m1 =$

How do you solve the initial value with 4y4yy0 y01 y015Solut

Solution

Get Help Now

Submit a Take Down Notice