Solve the initial value problem for the heat equation with time-dependent sources au = k+ Q(x, t) at ax u(x, 0) = f(x) subject to the following boundary conditions: Pls show the all steps, solve this thx (а) и(0, г) — А #0, u(L, t) = 0

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Q: b) Find the temperature u(x,t) from the heat equation in a laterally insulated copper bar 80 cm long…

A: b) Given :-  The temperature ux,t from the heat equation in a laterally insulated copper bar 80 cm…

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Q: Apply the initial conditions y(0) = 1, y'(0) = 4 to find c1 and c2 given that y = c1x + c2e2x

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Q: 4. y' +y = -x/y

A: As per our guideline we are supposed to answer only one asked question.kindly repost other question.…

Q: Find a function y=P(t)that is a solution of y'=.02y with initial condition y=100 when t=0.

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Q: 2²y" - 4.xy+6y= 0, y(2)=0, y'(2) = 4 %3D

A: The differential equation is given as x2y''-4xy'+6y=0x2D2y-4xDy+6y=0x2D2-4xD+6y=0 It is a Cauchy…

Q: homogeneous solution to the second order constant coefficient ODE’s. Solution must be real.

A: d3dx3y-3d2dx2y+3ddxy-y=0 A linear homogeneous ODE with constant coefficients has the form of…

Q: 2 A 3 cm length silver bar with a constant cross section area 1 cm² (density 10 g/cm³, thermal…

A: Let L, K, σ and ρ denote the length, thermal conductivity, specific heat and density, respectively.…

Q: Verify that the function u=e(-α^2)t sin x is a solution of the heat conduction equation ut = α2uxx

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Q: Consider the initial value problem (IVP): x²y" + xy'-y= 0, y(1)=0, y'(1)=2. It can be shown that y₁…

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Q: [23] R and f 5 sec(t) 0

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Q: y = C1cos(x) + C2sin(x) is a two-parameter of the: y" + y = 0. Look for a solution that is competes…

A: Here we put the initial conditions to find C1 and C2

Q: Solve the initial value problem: x²y"-2xy' +2y=0, y(1)=2, y'(1)=0.

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Q: Q2: (a): Solve (i) (x² D² – xD – 151I)y = 0 y(1) = 0.1, y'(1) = -4.5 (ii) 100 – 20 + d? N dN dt? dt

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Q: y? when y dy and x = 1. Find the particular solution to dz

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Q: 4. Solve the IVP y' = 2x-3y +e¯* y'= 2x- 3 y +e * with y(0) = 1 y(0)= 1

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Q: y" – y' = sin(t) y(0) = 1 y' (0) = -1

A: We have to solve the IVP           y''-y'=sinty0=1y'0=-1