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

Expert Solution

milcah answered 2 years ago

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.

Solve the given initial-value problem. y'' + 4y' + 4y = (5 + x)e^(−2x) y(0) =...

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.

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 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

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