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Popular >

y^'=(t+y)2-1,y(1)=6

Solution

y^{^{'}} = (t + y) 2 - 1, y (1) = 6

Solution

y = - t + 7 e^{2 t - 2}

Solution steps

y^{'} (t) = (t + y) 2 - 1

Solve linear ODE:

y = - t + 7 e^{2 t - 2}

y = - t + 7 e^{2 t - 2}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=(t+y)2-1,y(1)=6 ?
The solution for y^'=(t+y)2-1,y(1)=6 is y=-t+7e^{2t-2}