Reduce the system of first-order equations given below to a single equation of h

Question

Reduce the system of first-order equations given below to a single equation of higher order, and solve that equation.
x'(t)-2y(t)=t^2
y'(t)-x(t)+y(t)=1

Explanation / Answer

t = y^(1/3) x = y^(2/3) Or y = ±x^(3/2) y = y1(t) + y2(t) y = 1 + t + e^t y ' = 1 + e^t y '' = e^t substituting in original eqn t e^t - (1+t)(1 + e^t) + 1 + t + e^t = te^t - 1 - t - e^t - te^t + 1 + t + e^t = 0 so y1 and y2 satisfy the corresponding homogeneous eqn let yp = (At^2 + Bt + C)e^(2t) y ' (p) = 2At + B)e^(2t) + 2e^(2t)(At^2 + Bt + C) = e^(2t) [ 2At + B + 2At^2 + 2Bt + 2C) y '' p = e^(2t)[2A + 4At + 2B ] + 2e^(2t) [2At^2 + 2At + 2Bt + B + 2C ] = e^(2t) [ 2A + 4At + 2B + 4At^2 + 4At + 4Bt + 2B + 4C ] substituting in original eqn te^(2t) [ 2A + 4At + 2B + 4At^2 + 4At + 4Bt + 2B + 4C ] - (1+t)e^(2t) [ 2At + B + 2At^2 + 2Bt + 2C) + (At^2 + Bt + C)e^(2t) = t^2 e^(2t) =>t [ 2A+4At+ 2B+4At^2+ 4At+ 4Bt+ 2B+ 4C ]- (1+t)[ 2At+ B+ 2At^2+2Bt+ 2C)+(At^2+ Bt+ C)= t^2 => t^3(4A-2A) + t^2(4A+4A+4B-2A-2B-2A+A) + t(2A+2B+2B+4C-2A-2B-B-2C+B)-B-2C+C = t^2 => t^3(2A) + t^2(5A+ 2B) + t(2B+2C) - B- C = t^2 2A = 0 ==> A = 0 5A + 2B = 1 ==> B = 1/2 2B + 2C = 0 ==> 2C = -1 ==> C = -1/2 so y = y1 + y2 + yp y = 1 + t + e^t + (1/2)t e^(2t) - 1/2e^(2t)

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