What is the integral of e^{(x-y)} ?

The integral of e^{(x-y)} is -e^{x-y}+C

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integral of e^{(x-y)}

\int e^{(x - y)} d y

- e^{x - y} + C

Solution steps

\int e^{x - y} d y

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - e^{u}

Substitute back

u = x - y

= - e^{x - y}

Add a constant to the solution

= - e^{x - y} + C

\int t \cdot sec^{2} (2 t) d t

y^{^{'}} - \frac{y}{t} = 0

\int \frac{cos ^{3} ( 4 x )}{sin ( 4 x )} d x

t^{3} + 2 t^{2} - t + 4

3 x^{2} y^{2} d x + (2 y^{3} + 4 y^{3}) d y = 0