# Please solve the following differential equation: D^2y = 3sinx - 4y?

given y=0 at x=0 and dy/dx=1 at x=π/2

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So you mean, I think: d²y/dt² + 4y = 3sinx

For the solution of the homogenous equation d²y/dt² + 4y = 0 try : y = ℮^(m.x) then from the homogeneous differential equation we get (m²+4 ).℮^(mx) =0 which gives the auxiliary equation:

(m-j2).(m+j2) = 0

or m₁ = j2 and m₂ = -j.2 and the solution is y = A.cos(2x) + B.sin(2x)

We now need a particular integral or particular solution of the complete (non-homogeneous) equation. Try y=sinx , then the LHS: y" + 4y = -sinx +4.sinx = 3sinx so the solution fits! giving the general solution as:

y = A.cos(2x) + B.sin(2x) + sinx

The first boundary condition gives A=0

The second 2B.cos(2x) +cosx = 1 or -2B = 1 or B=-½ giving the solution as:

y = sin(x) - ½.sin(2x)

Maths is beautiful!

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• so which you will possibly be able to desire to locate the polynomials and their roots and stuff. a) The polynomial is x^2+3x-4 = 0 and the roots are 4 and -a million. those men flow in as alpha and beta interior the final answer y = Ae^(alpha x) + Be^(beta x) So the answer is y = Ae^(4x) + B e ^(-a million) b) The polynomial is x^2-4x +5 = 0 which has the roots (a million/2)(4 +/- sqrt(sixteen - 20)) = (a million/2)(4 +/- 2i) = 2 +/- i. So the answer is y = Ae^(2+i) + Be^(2-i) c) The polynomial is x^2+2x+a million = (x+a million)^2 The repeated root ability we take as our answer foundation { e^-x, xe^-x} so as that a conventional answer Ae^-x + Bx e^-x.

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• math sucks

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