T F If y' = 3y^(2) + 5y − 2 and y(1) = 0, then lim t→∞ y(t) = −2.

Expert Solution

answer is TRUE

Colby Messinger answered 2 years ago

Use the Laplace transform to solve the IVP: y^'''+y^''+3y^'-5y =16e^(-t); y(0)=0; y'(0)=2; y^'' (0)= -4

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1...

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1 given X is the number of students who get an A on test 1 given Y is the number of students who get an A on test 2 find the probability that more then 90% students got an A test 2 given that 85 % got an A on test 1

Solve the initial value problem: Y''-4y'+4y=f(t) y(0)=-2, y'(0)=1 where f(t) { t if 0<=t<3 , t+2...

Solve the initial value problem: Y''-4y'+4y=f(t) y(0)=-2, y'(0)=1 where f(t) { t if 0<=t<3 , t+2 if t>=3 }

y"-3y'+2y=4t-8 , y(0)=2 , y'(0)=7 y(t)=?

ﬁnd the general solution of the given differential equation. 1. y''+2y'−3y=0 2. 6y''−y'−y=0 3. y''+5y' =0 4. y''−9y'+9y=0

use laplace to answer y"-3y'+2y=1+cost+e^-t,y(0)=1,y'(0)=0

Solve the equation below for y(t): y''+2y'-3y=8u(t-3): y(0) = 0; y'(0)=0

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0)...

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0) = 1, solution: y(t) = et2+2t y′ = 5t4y, y(0) = 1, solution: y(t) = et5 y′ = t3/y2, y(0) = 1, solution: y(t) = (3t4/4 + 1)1/3 For the IVPs above, make a log-log plot of the error of Backward Euler and Implicit Trapezoidal Method, at t = 1 as a function of hwithh=0.1×2−k for0≤k≤5.

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0)...

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0) = 1, solution: y(t) = et2+2t y′ = 5t4y, y(0) = 1, solution: y(t) = et5 y′ = t3/y2, y(0) = 1, solution: y(t) = (3t4/4 + 1)1/3 For the IVPs above, make a log-log plot of the error of Explicit Trapezoidal Rule at t = 1 as a function ofhwithh=0.1×2−k for0≤k≤5.

Solve the differential equation by Laplace transform y^(,,) (t)-2y^' (t)-3y(t)=sint where y^' (0)=0 ,y=(0)=0

Question

T F If y' = 3y^(2) + 5y − 2 and y(1) = 0, then lim t→∞ y(t) = −2.

Solutions

Expert Solution

Related Solutions

Use the Laplace transform to solve the IVP: y^'''+y^''+3y^'-5y =16e^(-t); y(0)=0; y'(0)=2; y^'' (0)= -4

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1...

Solve the initial value problem: Y''-4y'+4y=f(t) y(0)=-2, y'(0)=1 where f(t) { t if 0<=t<3 , t+2...

y"-3y'+2y=4t-8 , y(0)=2 , y'(0)=7 y(t)=?

ﬁnd the general solution of the given differential equation. 1. y''+2y'−3y=0 2. 6y''−y'−y=0 3. y''+5y' =0 4. y''−9y'+9y=0

use laplace to answer y"-3y'+2y=1+cost+e^-t,y(0)=1,y'(0)=0

Solve the equation below for y(t): y''+2y'-3y=8u(t-3): y(0) = 0; y'(0)=0

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0)...

y′ = t, y(0) = 1, solution: y(t) = 1+t2/2 y′ = 2(t + 1)y, y(0)...

Solve the differential equation by Laplace transform y^(,,) (t)-2y^' (t)-3y(t)=sint where y^' (0)=0 ,y=(0)=0