Consider the system of higher order differential equations y″ = t¯¹y' + 7y – tz + (sint)z' + e5t, z" = y — 3z'. Rewrite the given system of two second order differential equations as a system of four first order linear differential equations of the form ÿ' = P(t)ÿ+ g(t). Use the following change of variables y' [Y] Y1 Y2 Y3 LY4_ help (formulas) help (matrices) Book: Section 3.3 of Notes on Diffy Qs [y1(t)] [ y(t)] Y2(t) y' (t) ÿ(t) = = Y3(t) z(t) Y₁(t)] [z'(t)]

Find a differential equation for y=c1e^9t +c2e^-4t. O y"+5y +16y=0 O y"-5y'-16y=0 Oy"+5y'+36y=0 Oy"-5y'-36y=0

Transform the given differential equation into an ferential equations. equivalent system of first-order dif- 31" = 2y', and 3" = -2r'

Convert the differential equation u''+5u'−3u=e−4tu′′+5u′-3u=e-4t into a system of first order equations by letting x=u, y=u'x=u, y=u′x'x′ =   y'y′ =

we are given a second-order non-homogeneous linear differential equation: mh''+5h'+120h=mg with initial conditions h(0)=100 and h'(0)=0.Converting into a system of first-order differential equations. We can do this by letting v = h' (velocity is the derivative of height with respect to time). Then, the original equation becomes mv' + 5v + 120h = mg, which is a first-order differential equation in v. We also have the equation h' = v, which is a first-order differential equation in h. Write a python code to solve the system for h(t) and v(t), where g is 9.8 and m varies.

Which of the following systems of differential equations would dictate the rates of change of the salt concentration for the two-tank system shown below? 10 L/min 70 g/L VAL 14 L/min I 6 L/min OC A/= (1/VA) (10 × 70 + 6cg - 14CA), c_l= (1/V₂) (14c. – 16c) OC Al= (1/VA) (70 – 6cв - 14CA), CB= (1/VB) (14CA - 16CB) 16св) VB L C Al= (1/VA) (14CA CB/= (1/VB) (14CA - 16CB) 10 × 70 – 6cß), C Al= (1/VA) (10 × 70+ 16cB - 14CA), CB/= (1/VB) (14CA 10CB) 10 L/min

3t Convert the differential equation u"+u'-u = e³t into a system of first order equations by letting x = u, y = u' x' = y' =

Solve the system [: 16 -36 dx dt 6. -14 with the initial value 13 x(0) 5 -5e^(4t)+13e^(-3t) x(t) = -5e^(4)+10e^(-3t)