Question

In: Math

solve the first order equation with initial value y’ +3/4y = x^6,; y(1) =2

solve the first order equation with initial value y’ +3/4y = x^6,; y(1) =2

Solutions

Expert Solution


Related Solutions

Solve the initial value problem: y'' + 4y' + 4y = 0; y(0) = 1, y'(0)...
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.
a) Solve the Cauchy-Euler equation: x^2y'' - xy' + y = x^3 b) Solve the initial-value...
a) Solve the Cauchy-Euler equation: x^2y'' - xy' + y = x^3 b) Solve the initial-value problem: y'' + y = sec^3(x); y(0) = 1, y'(0) =1/2
Solve the given initial-value problem. y'' + 4y = −1,    y(π/8) =3/4,y'(π/8) = 2 ,
Solve the given initial-value problem. y'' + 4y = −1,    y(π/8) =3/4,y'(π/8) = 2 ,
Solve the initial value problem. (x^2 * D^2 +xD - 4i) * y = x^3, y(1)...
Solve the initial value problem. (x^2 * D^2 +xD - 4i) * y = x^3, y(1) = -4/5, y'(1) = 93/5
Solve the following initial value: y ''+ 4y = 2 cos 2t, y(0) = −2 and...
Solve the following initial value: y ''+ 4y = 2 cos 2t, y(0) = −2 and y 0 (0) = 0
A) Solve the initial value problem: 8x−4y√(x^2+1) * dy/dx=0 y(0)=−8 y(x)= B)  Find the function y=y(x) (for...
A) Solve the initial value problem: 8x−4y√(x^2+1) * dy/dx=0 y(0)=−8 y(x)= B)  Find the function y=y(x) (for x>0 ) which satisfies the separable differential equation dy/dx=(10+16x)/xy^2 ; x>0 with the initial condition y(1)=2 y= C) Find the solution to the differential equation dy/dt=0.2(y−150) if y=30 when t=0 y=
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 }
(3 pts) Solve the initial value problem 25y′′−20y′+4y=0, y(5)=0, y′(5)=−e2. (3 pts) Solve the initial value...
(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 following initial value problem: y'''-4y''+20y'=-102e^3x, y(0)=3, y'(0)=-2, y''(0)=-2
Solve the following initial value problem: y'''-4y''+20y'=-102e^3x, y(0)=3, y'(0)=-2, y''(0)=-2
(x+3)^2/3 + (x+3)^1/3 - 6 = 0 Solve the equation.
(x+3)^2/3 + (x+3)^1/3 - 6 = 0 Solve the equation.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT