Given y1 (t) = t and y2(t) = t¯1 satisfy the corresponding homogeneous equation of t²y' – 2y = - 1+ 2t, t>0 Then the general solution to the non-homogeneous equation can be written as y(t) = ci¥ı(t) + c2y2(t) +Y(t). Use variation of parameters to find Y(t). Y(t) = Preview %3D

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Put the equation in standard form of linear equation and find the solution

ts) Find two linearly independent solutions of 2x²y" - xy + (2x + 1)y=0, x > 0 of the form Y₁ = x¹(1+ a₁ + a₂c² + a3x³ + ...) Y₂ = x²(1 + b₁c + b₂x² + b3x³ + ...) where T1 T2. Enter T1 T2 a1 = a₂ = a3 = b₁ b₂ TE b3 N - || || || || =

Eliminate the arbitrary constant. -X 2x y=c₂₁e + c₂e²t 1 A B C D y"-y'- 2x=0 y"-y'-2y=0 2y "'-y'- 2x=0 y"-2y'- 2x=0

check whether or not the given equations are homogeneous of some degree, and if yes, use an appropriate variable change to solve them

The position of a particle moving along the y-axis is given by y(t) = ct³ + bt + 100, where c and b are constants, y(t) is measured in feet, and t is measured in minutes. If the velocity of the particle is 350 feet per minute at t = 5 and the acceleration of the particle is 120 feet per minute per minute at t = 5, what is the value of b?

DETERMINE SOLUTION OF DE 0 + x X Ⓒ) (2x + 3x - 13) dx + 3x - 2y) dy = 0 dx + y = x2

Let us solve the following equation via separation of variables J²u(x, t) Ət² = 3. a²u(x, t) ?х2 a) After using the ansatz u(x, t) = X(x)T(t), which equations do you get? Pick only two. (below C' denotes a constant) Ə²X(1) J²X(z) a²T(t) 842 √3CT(t) a²T(t) 84² = 1 T(t) 8²X(z) Əx² √C X(T) = C X(T) 8x² Ər² a²T(t) 3C T(t) b) We are interested in the oscillatory solution given by X(x) = a₁ cos ( $(√|C|x) + a2 sin (√√C\x) If the boundary conditions are u(x = 0,t) = 0 u(x = 1,t) = 0 Identify the correct form of the solution: O X(x) = a₂ sin(n+x) O X(x) = a₂ cos(NTT) O X(z) = a2 sin((2n-1)) = = = X(T) = OX(x) = a₂ cos((2n-1) -x)

I. Differentiate the following (show your solution): 1. y = 1+3x-x² 2. y=2 x - x+ 3 x'+x?+5 х+ 3х +x* + 5 ²- 2 x² - 3 3. y=