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Popular Calculus >

y^'=7x+y,y(0)=5

Solution

y^{^{'}} = 7 x + y, y (0) = 5

Solution

y = - 7 (x + 1) + 12 e^{x}

Solution steps

y^{'} (x) = 7 x + y

Solve linear ODE:

y = - 7 (x + 1) + 12 e^{x}

y = - 7 (x + 1) + 12 e^{x}

Graph

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\int_{0}^{π} 12 sin^{4} (x) d x

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\int_{1}^{2} (7 x + x^{2}) d x

derivative of cos(2^x)

d er i v a t i v e cos (2^{x})

Frequently Asked Questions (FAQ)

What is the solution for y^'=7x+y,y(0)=5 ?
The solution for y^'=7x+y,y(0)=5 is y=-7(x+1)+12e^x