(3 pts) Solve the initial value problem
25y′′−20y′+4y=0, y(5)=0, y′(5)=−e2.
(3 pts) Solve the initial value problem
y′′ − 2√2y′ + 2y = 0, y(√2) = e2, y′(√2) = 2√2e2.
Consider the second order linear equation t2y′′+2ty′−2y=0,
t>0.
(a) (1 pt) Show that y1(t) = t−2 is a solution.
(b) (3 pt) Use the variation of parameters method to obtain a
second solution and a general solution.
solve the initial values:
if Y(3)-4Y"+20Y'=51e^3x
Y"(0)=41, Y'(0)= 11. Y(0)= 7 > solution is Y(x)= e^3x+2 e^2x
sin(4x)+6
so, what is the solution for:
Y(3)-8Y"+17Y'=12e^3x
Y"(0)=26, Y'(0)= 7. Y(0)= 6
Y(x)=???
Solve the initial value problem: y'' + 4y' + 4y = 0; y(0) = 1,
y'(0) = 0.
Solve without the Laplace Transform, first, and then with the
Laplace Transform.
Solve the given initial-value problem. y'' + 4y' + 4y = (5 +
x)e^(−2x) y(0) = 3, y'(0) = 6
Arrived at answer
y(x)=3e^{-2x}+12xe^{-2x}+(15/2}x^2e^{-2x}+(5/6)x^3e^{-2x) by using
variation of parameters but it was incorrect.
Consider the following initial value problem.
y''−4y = 0,
y(0) = 0, y'(0) = 5
(a) Solve the IVP using the characteristic equation method from
chapter 4.
(b) Solve the IVP using the Laplace transform method from chapter
7.
(Hint: If you don’t have the same final answer for each part, you’ve
done something wrong.)