Find the general solution of y'' + 2(sech^2 t)y = 0 (1), given that y1 =...

Find the general solution of y'' + 2(sech^2 t)y = 0 (1), given that y1 = tanh t is a solution to (1)

Expert Solution

milcah answered 2 years ago

y1(t) = (1+t)2 is a solution to y'' + p(t)y' + q(t)y = 0. Find a...

y1(t) = (1+t)2 is a solution to y'' + p(t)y' + q(t)y = 0. Find a second solution that is linearly indepentent of y1(t).

x2 y" + (x2+x) y’ +(2x-1) y = 0, Find the general solution of y1 with...

x2 y" + (x2+x) y’ +(2x-1) y = 0, Find the general solution of y1 with r1 and calculate the coefficient up to c4 and also find the general expression of the recursion formula, (recursion formula for y1) Find the general solution of y2 based on theorem 4.3.1. (Hint, set d2 = 0)

1) find the solution t the non-homogenous DE y''-16y=3e5x , y(0)=1 , y'(0)=2 2)find the solution...

1) find the solution t the non-homogenous DE y''-16y=3e5x , y(0)=1 , y'(0)=2 2)find the solution to the DE using cauchy-euler method x2y''+7xy'+9y=0 , y(1)=2 , y'(1)=3 3)find the solution to the DE using Laplace y''+8y'+16y=0 , y(0)=-1 , y'(0)=8

Find a general solution to y” + y = (tan t)^2

find the general solution of the given differential equation. 1. y'' + y = tan t,...

find the general solution of the given differential equation. 1. y'' + y = tan t, 0 < t < π/2 2. y'' + 4y' + 4y = t-2 e-2t , t > 0 find the solution of the given initial value problem. 3. y'' + y' − 2y = 2t, y(0) = 0, y'(0) = 1

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.

Question

Find the general solution of y'' + 2(sech^2 t)y = 0 (1), given that y1 =...

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