Question

Having some trouble finding c1 and c2,

Solve the IVP

y'' - 4y' + 13y = 0, y(pi) = 3, y(pi) = 2

Thank you!!

Answer #1

solve y''+4y'+4y=0 y(0)=1, y'(0)=4 IVP

The general solution of the equation
y′′+6y′+13y=0 is
y=c1e-3xcos(2x)+c2e−3xsin(2x)
Find values of c1 and c2 so that y(0)=1
and y′(0)=−9.
c1=?
c2=?
Plug these values into the general solution to obtain
the unique solution.
y=?

Solve the IVP y¨ − 5 y˙ + 4y = e^t , y(0) = 3, y˙(0) = 1/3

solve the IVP
y'' - 4y' - 5y = 6e-x, y(0)= 1, y'(0) =
-2

Use the definition of the Laplace transform to solve the
IVP:
4y''− 4y' + 5y = δ(t), y(0) = −1, y'(0) = 0.

Use python code to solve IVP. Code must print both general and
particular solutions. Please paste full python code for your
answer. IVP: y′=(5x^2)(y^2)+(125x^2), y(−7π)=10
Below is an example of code for the problem 7x″−3x′−57x=0,
x(2π)=9, x′(−π)=5exp(3) :
from sympy import *
y = symbols('y')
x = Function('x')
x = dsolve(7*diff(x(y),y,2)-3*diff(x(y),y)-57*x(y),x(y))
print('The general solution is: ',x,'.',sep='')
x = x.rhs
xp = diff(x,y)
y0, y0p = 2*pi, -pi
x0, x0p = 9, 5*exp(3)
x0LHS, x0pLHS = x.subs(y,y0) - x0, xp.subs(y,y0p) -...

Solve the IVP with Cauchy-Euler ODE: x^2 y''+3xy'+4y=0; y(1)=0,
y’(1)=−2

Question : y'' = 4y' + 13y =0 , (y''+ 2y' + 2y)^2 = 0 , y'' - y'
- 2y = cosx , y'' + 3y' + 2y = x^2 - e^2x

y'' + 4y' + 5y = δ(t − 2π),
y(0) = 0, y'(0) = 0
Solve the given IVP using the Laplace Transform. any help
greatly appreciated :)

Solve the following second-order equation applying variation of
parameters method:
y'' + 4y' + 4y = t^(-2) * e^(-2t) t > 0
Thank you!

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