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Popular >

y^'+y=e^x,y(0)=1

Solution

y^{^{'}} + y = e^{x}, y (0) = 1

Solution

y = \frac{e ^{x}}{2} + \frac{1}{2 e ^{x}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x}}{2} + \frac{1}{2 e ^{x}}

y = \frac{e ^{x}}{2} + \frac{1}{2 e ^{x}}

Popular Examples

tangent of e^x+1 tangent of

e^{x} + 1

(e^{5 x})^{^{'}}

limit as x approaches 2+of (x-3)^2

x \to 2 + lim ((x - 3)^{2})

integral of (x^2-5x+6)/((2x-1)(x-2)^2)

\int \frac{x ^{2} - 5 x + 6}{( 2 x - 1 ) ( x - 2 ) ^{2}} d x

limit as x approaches infinity of x^3-3

x \to \infty lim (x^{3} - 3)

Frequently Asked Questions (FAQ)

What is the solution for y^'+y=e^x,y(0)=1 ?
The solution for y^'+y=e^x,y(0)=1 is y=(e^x)/2+1/(2e^x)