This is the fourth part of a four-part problem.If the given solutionsÿ₁ (t) =2e³t 8e3est 20e-tform a fundamental set (i.e., linearly independent set) of solutions for the initial value problemÿ' =2 (t) =ÿ, 7(0) ==2e³8e-3e320e-t8e³+2e712e³t+5e7[17],impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks.8e³+2e--O [12³ + 5e-²g(t)=

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Answered: If the given solutions ÿ₁ (t) =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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d 39. (a) Show that a one-parameter family of solutions of the equation (b) (4xy + 3x²) dx + (2y + 2x²) dy = 0 is x² + 2x²y + y² = c. Show that the initial conditions y(0) = -2 and y(1) = 1 determine the same implicit solution. (c) Find explicit solutions y₁(x) and y2(x) of the differential equation in part (a) such that y₁ (0) = -2 and y₂(1) = 1.
If the given solutions 2 – 2t yi(t): y2(t) = 2t form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem 21-2 1– 21-1 + 21-2 -23 y, y(3) = t > 0, y -2t-2 2t-1 – 21-2 -34 impose the given initial condition and find the unique solution to the initial value problem for t > 0. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks. 2t 0) = At) = ( + ( 2t
1) Find the solution to each equation with the following. Initial conditions: part a) dy. 11+ 4 = 1 Y(2)=1 part b) dy 21:16 Yt+2t 4(-1)=1 part c) yox dy + x = 5 Y(5)=1
Thus far in the class we've only seen one kind of so-called "linear" difference equation models: the Malthusian ones of the form N(t+1)= XN (t) This is called linear because N(t+1) is a linear function of N(t) - in other words, if we view N(t+1) as "y" and N(t) as "r" then this is an equation of the form "y = mx" since À is a constant. Another more general type of "linear" difference equation is: N(t+1) = AN(t) + a where a is some constant. This is again a linear model because N(t+1) is a linear function of N(t), this time of the form "y=mx+b". This is sometimes called a "linear difference equation with a constant term". The next few problems have to do with linear difference equations with constant terms. 4. Consider the difference equation N(t + 1) = 0.5N(t) + 40 where we are given that N(0) = 100. a) Check that N(t) = 100* (0.5) is not a solution to this difference equation. b) Check that N(t) = 100* (0.5) +40 is not a solution to this difference equation. c) Check that N(t) = 20 *…
Two tanks T1 and T2, respectively contain 200 and 100 liters of water and unspecified amounts of fertilizer. Now, set up a model describing the amount of fertilizer contained in each tank y1(t) and y2(t) at time t. please do not provide solution in image format thank you!
Consider the initial-value problem y = x(y² - 1)¹/3 (y4 -8y)2/3, y(0) = yo, where yo is a real number. Which of the following values of yo does the problem have a unique solution? (i) -1 (ii) 0 (iii) 1 (iv) 2 (v) 3
. Solve the following initial value problem. -2 -2 6 -2 & y = ( ²2 √ ) Y, Y(0) = ( ² ) -Y dt -2 -1 -1
Question. Solve: x" -x + 5y² = + 4" - 4y - 2x² = -2 with x (0) = y(0) = x²(0) = y₁ (0) = 0 Let & [ X (²) } = U (6) & & Gy (+) 2 = V(c) Evaluate U(2) Evaluate (²) Evaluate x(1) Evaluate y(1)
(a) If b – 4ac = 0 and y1 = em# solves ay" + by' + cy = 0, use reduction of order to get a linearly independent solution. [You may assume linear independence]

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