Question

In: Advanced Math

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).

Solutions

Expert Solution

Colby Messinger answered 1 year ago
Next > < Previous

Related Solutions

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)
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
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.
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)
Let   y(t) = (1 + t)^2 solution of the differential equation y´´ (t) + p (t) y´...
Let   y(t) = (1 + t)^2 solution of the differential equation y´´ (t) + p (t) y´ (t) + q (t) y (t) = 0 (*) If the Wronskian of two solutions of (*) equals three. (a) ffind p(t) and q(t) (b) Solve y´´ (t) + p (t) y´ (t) + q (t) y (t) = 1 + t
if y(t) is the solution of y′′+2y′+y=δ(t−3),y(0)=0,y′(0)=0 the find y(4)
if y(t) is the solution of y′′+2y′+y=δ(t−3),y(0)=0,y′(0)=0 the find y(4)
1) Find y as a function of t if 9y′′+24y′+32y=0, y(0)=5,y′(0)=8. y(t)= 2) Find y as...
1) Find y as a function of t if 9y′′+24y′+32y=0, y(0)=5,y′(0)=8. y(t)= 2) Find y as a function of x if y′′′+16y′=0, y(0)=−5,  y′(0)=−32,  y′′(0)=−32. y(x)= 3) Find y as a function of t if 9y′′−12y′+40y=0, y(1)=5,y′(1)=9. y=
Let X = [1, 0, 2, 0]tand Y = [1, −1, 0, 2]t. (a) Find a...
Let X = [1, 0, 2, 0]tand Y = [1, −1, 0, 2]t. (a) Find a system of two equations in four unknowns whose solution set is spanned by X and Y. (b) Find a system of three equations in four unknowns whose solution set is spanned by X and Y. (c) Find a system of four equations in four unknowns that has the set of vectors of the form Z + aX + bY as its general solution where...
1. solve the initial value problem. (t^(2)+1)y'+2ty=tant , y(0)=2 2.find the solution to this initial value...
1. solve the initial value problem. (t^(2)+1)y'+2ty=tant , y(0)=2 2.find the solution to this initial value problem. yy'=e^x+x , y(0)=y_0 y_0 is a nonzero constant.
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT