solve using elimination method: 2x\'+y\'-y=t, x\'+y\'=t^2, x(0)=1, y(0)=0 Soluti

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2x'+y'-y = t x'+(x'+y')-y = t => x'+t^2-y = t => t^2-y' +t^2-y = t => y'+y = 2t^2-t => dy/dt + y = 2t^2-t IF = e^integral (1) dt = e^t y e^t = integral ( 2t^2-t) e^t dt = (2t^2-t) e^t - integral (4t-1) e^t dt = (2t^2-t) e^t -(4t-1) e^t + integral (4) e^t dt ye^t = (2t^2-t) e^t -(4t-1) e^t + 4 e^t + C => y = (2t^2-t) -(4t-1) +4 = 2t^2-5t+5 + C => y' = 4t-5 x'+y' = t^2 x' = t^2-y' x' = t^2-4t+5 dx = (t^2-4t+5 ) dt integrate x =(1/3)t^3 -2t^2+5t + C x(0) = 1 => C = 1 hence x(t) = (1/3)t^3 -2t^2+5t + 1 y(0) = 0 => C = -5 hence y(t) = 2t^2-5t

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