What is the integral of 2ye^{2x} ?

The integral of 2ye^{2x} is ye^{2x}+C

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integral of 2ye^{2x}

\int 2 y e^{2 x} d x

y e^{2 x} + C

Solution steps

\int 2 y e^{2 x} d x

Take the constant out:

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

= 2 y \cdot \int e^{2 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= 2 y \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= 2 y \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 2 y \frac{1}{2} e^{u}

Substitute back

u = 2 x

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

Simplify

2 y \frac{1}{2} e^{2 x} : y e^{2 x}

= y e^{2 x}

Add a constant to the solution

= y e^{2 x} + C

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 1}

\frac{d}{d t} (tan (t + \frac{π}{4}))

\frac{d y}{d x} = y - y^{5}

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} + a y = 0, a > 1

\frac{x ^{5} + 1}{x}