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).

Expert Solution

Colby Messinger answered 1 year ago

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)

Consider the differential equation, L[y] = y'' + p(t)y' + q(t)y = 0, (1) whose coefficients...

Consider the differential equation, L[y] = y'' + p(t)y' + q(t)y = 0, (1) whose coefficients p and q are continuous on some open interval I. Choose some point t0 in I. Let y1 be the solution of equation (1) that also satisfies the initial conditions y(t0) = 1, y'(t0) = 0, and let y2 be the solution of equation (1) that satisfies the initial conditions y(t0) = 0, y'(t0) = 1. Then y1 and y2 form a fundamental set...

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=

given y1 find y2 (Differential Equations) (3x-1)y''-(3x+2)y'-(6x-8)y=0 and y1=e^(2x)

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

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Related Solutions

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

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

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)...

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

Let y(t) = (1 + t)^2 solution of the differential equation y´´ (t) + p (t) y´...

if y(t) is the solution of y′′+2y′+y=δ(t−3),y(0)=0,y′(0)=0 the find y(4)

Consider the differential equation, L[y] = y'' + p(t)y' + q(t)y = 0, (1) whose coefficients...

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...

given y1 find y2 (Differential Equations) (3x-1)y''-(3x+2)y'-(6x-8)y=0 and y1=e^(2x)