Write the given third order linear equation as an equivalent system of first order equations with initial values. Use Y1 = Y, Y2 = y', and y3 = y". - - √ (3t1 + 3 − t3)y" — y" + (3t2 + 3)y' + (3t — 3t1) y = 1 − 3t2 \y(3) = 1, y′(3) = −2, y′′(3) = −3 (8) - (888) - with initial values Y = If you don't get this in 3 tries, you can get a hint.

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