What is the integral of e^{ty} ?

The integral of e^{ty} is 1/t e^{ty}+C

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integral of e^{ty}

\int e^{t y} d y

\frac{1}{t} e^{t y} + C

Solution steps

\int e^{t y} d y

Apply u-substitution

: \int \frac{e ^{u}}{t} d u

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

Take the constant out:

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

= \frac{1}{t} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{t} e^{u}

Substitute back

u = t y

= \frac{1}{t} e^{t y}

Add a constant to the solution

= \frac{1}{t} e^{t y} + C

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = 0

\int \frac{x}{x + 7} d x

y^{^{''}} + 4 y = 4 sec (2 t)

x y^{^{'}} - y = 2020 x^{2}

(- 7, - 4), (1, - 6)