Question

Solve the given initial-value problem. 5y'' + y' = −6x, y(0) = 0, y'(0) = −25

The answer I got is 155-155e^(-1/5)x-3x^2-30x

Answer #1

we first find auxilary
equation then its roots to find y_{c}.to find y_{p}
we use method of undetermined coefficients.to find constant we
apply initial condition.in y_{p} there is a constant term c
but writing y(x) we can added this to constant term c1 in
y_{c}.so we get a constant
c_{3}=c+c_{1}.

I think there is a problem in your answer.

My answer is

y(x)=-3x^{2}+30x-275+275e^{(-x/5)}

solve the given initial value problem y''-5y'+6y=0, y(0)=3/5,
y'(0)=1

Solve the given initial-value problem.
y'' + 5y' −
6y =
12e2x, y(0)
= 1, y'(0) = 1

Use Laplace transforms to solve the given initial value
problem.
y"-2y'+5y=1+t y(0)=0 y’(0)=4

Solve the initial value problem
x′=x+y
y′= 6x+ 2y+et,
x(0) = 0, y(0) = 0.

solve the given initial value problem
dx/dt=7x+y x(0)=1
dt/dt=-6x+2y y(0)=0
the solution is x(t)= and y(t)=

solve the initial value problem y''-2y'+5y=u(t-2) y(0)=0
y'(0)=0

Use the simplex method to solve the linear programming
problem.
Maximize
P = 6x + 5y
subject to
3x
+
6y
≤
42
x
+
y
≤
8
2x
+
y
≤
12
x ≥ 0, y ≥ 0
The maximum is P =
at
(x, y) =

Solve the following initial value problem. y(4) − 6y′′′ + 5y′′
= 2x, y(0) = 0, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
(not using Laplace)

Find the solution of the initial value problem y′′+5y′+6y=0,
y(0)=11 and y′(0)=-25

Solve the initial-value problem y''+5y'+6y=0, y(0)=1,
y′(0)=−1,
using Laplace transforms. When writing your answer limit
yourself to showing:
(i) The equation for L(y);
(ii) The partial fraction decomposition;
(iii) The antitransforms that finish the problem.

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