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Popular >

y^'=8x+y,y(0)=4

Solution

y^{^{'}} = 8 x + y, y (0) = 4

Solution

y = - 8 (x + 1) + 12 e^{x}

Solution steps

y^{'} (x) = 8 x + y

Solve linear ODE:

y = - 8 (x + 1) + 12 e^{x}

y = - 8 (x + 1) + 12 e^{x}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=8x+y,y(0)=4 ?
The solution for y^'=8x+y,y(0)=4 is y=-8(x+1)+12e^x