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Popular >

y^'=4x+y,y(0)=5

Solution

y^{^{'}} = 4 x + y, y (0) = 5

Solution

y = - 4 (x + 1) + 9 e^{x}

Solution steps

y^{'} (x) = 4 x + y

Solve linear ODE:

y = - 4 (x + 1) + 9 e^{x}

y = - 4 (x + 1) + 9 e^{x}

Graph

Popular Examples

tangent of y=-3x^2,(-4,-48)tangent of

y = - 3 x^{2}, (- 4, - 48)

integral from 0 to 1 of te^t

\int_{0}^{1} t e^{t} d t

integral from 9 to 10 of-6

\int_{9}^{10} - 6 d x

integral from 1 to 4 of 1/x

\int_{1}^{4} \frac{1}{x} d x

integral of (x^2-x+10)/(x^3+5x)

\int \frac{x ^{2} - x + 10}{x ^{3} + 5 x} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=4x+y,y(0)=5 ?
The solution for y^'=4x+y,y(0)=5 is y=-4(x+1)+9e^x