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Popular >

xy^'+y=e^x,y(1)=2

Solution

x y^{^{'}} + y = e^{x}, y (1) = 2

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

derivative of sqrt(cos^2(x-sin^2(x)))

\frac{d}{d x} (cos^{2} (x) - sin^{2} (x))

derivative of arctan((x+1/(x-1)))

\frac{d}{d x} (arctan (\frac{x + 1}{x - 1}))

integral of 1/((x+1)^3(x+2))

\int \frac{1}{( x + 1 ) ^{3} ( x + 2 )} d x

derivative of f(x)=(2x-2)^4(x^2+x+1)^5 derivative of

f (x) = (2 x - 2)^{4} (x^{2} + x + 1)^{5}

(\partial)/(\partial x)(5ye^x)

\frac{\partial}{\partial x} (5 y e^{x})

Frequently Asked Questions (FAQ)

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