sqrt(x)(dy)/(dx)=e^{7y+sqrt(x)}

⚠ sqrt(x)(dy)/(dx)=e^{7y+sqrt(x)} is read as sqrt(x)*y'=e^(7y+sqrt(x)): use dif(expression, variable) for a derivative with respect to that variable, use (numerator)/(denominator) for a quotient, use * for multiplication, use ^ for a power, use ^(...) to group an exponent

1) Solve the separable equation

\sqrt{x} \cdot y' = e^{7 y + \sqrt{x}}

, where

y

is a function of

x

Where

\sqrt{x} \neq 0

, divide by

\sqrt{x}

to isolate

y'

y' = \frac{e^{7 y + \sqrt{x}}}{\sqrt{x}}

⇥1.1) Let's evaluate the domain of

\frac{e^{\sqrt{x}}}{\sqrt{x}}

⇥ For the expression

\frac{e^{\sqrt{x}}}{\sqrt{x}}

to be defined, we must have:

x > 0

⇥Solution obtained:

x > 0

The function of

x

is defined for

x > 0

ⓘ The exponential of a sum is the product of the exponentials:

e^{7 y + \sqrt{x}} = e^{\sqrt{x}} e^{7 y}

The right side is the product of

\frac{e^{\sqrt{x}}}{\sqrt{x}}

, which depends on

x

only, and

e^{7 y}

, which depends on

y

only

e^{7 y}

is never zero: separate the variables,

\frac{dy}{e^{7 y}} = \frac{e^{\sqrt{x}}}{\sqrt{x}} \cdot dx

⇥1.2) Let's compute the integral:

\int \frac{1}{e^{7 y}} d y

⇥For the math expression:

\frac{1}{e^{7 y}}

, since (1/a^n)' = -(n*ln(a))'/a^n, we try differentiating

\frac{1}{e^{7 y}}

⇥1.3) Let's perform the differentiation:

\frac{d}{d y} (\frac{1}{e^{7 y}})

⇥For

p3

1

and

q4

e^{7 y}

, use the differentiation rule:

\frac{d}{d y} (\frac{p3}{q4}) = \frac{\frac{d}{d y} (p3) \cdot q4 - p3 \cdot \frac{d}{d y} (q4)}{{q4}^{2}}

⇥1.4) Let's perform the differentiation

p3

\frac{d}{d y} (1)

⇥ⓘ The derivative of a constant is zero:

\frac{d}{d y} (1)

0

⇥

0

⇥1.5) Let's differentiate

q4

\frac{d}{d y} (e^{7 y})

⇥Apply the differentiation rule:

\frac{d}{d y} (\exp (f)) = \frac{d}{d y} (f) \cdot \exp (f)

with f =

7 y

⇥ⓘ The derivative of exp(x) is exp(x):

\frac{d}{d y} (e^{7 y})

\frac{d}{d y} (7 y) \cdot e^{7 y}

⇥

\frac{d}{d y} (7 y) \cdot e^{7 y}

⇥ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d y} (7 y)

7

⇥

7 e^{7 y}

⇥1.6) Let's evaluate the resulting derivative (

\frac{p3}{q4}

)':

\frac{0 \cdot e^{7 y} - 1 \cdot 7 e^{7 y}}{{(e^{7 y})}^{2}}

⇥ⓘ A product with a zero factor is zero:

0 \cdot e^{7 y}

0

■A power of a power multiplies the exponents, (a^n)^m = a^(n*m):

{(e^{7 y})}^{2}

e^{7 y \cdot 2}

⇥

\frac{0 - 7 e^{7 y}}{e^{7 y \cdot 2}}

⇥ⓘ Adding 0 changes nothing:

0 - 7 e^{7 y}

- 7 e^{7 y}

⇥ⅰ

7 \cdot 2

14

⇥

- \frac{7 e^{7 y}}{e^{14 y}}

⇥ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

- \frac{7 e^{7 y}}{e^{14 y}}

-7 e^{7 y - 14 y}

⇥

-7 e^{7 y - 14 y}

⇥ⓘ Like terms are grouped by adding their coefficients:

7 y - 14 y

-7 y

⇥

- 7 e^{- 7 y}

⇥ⓘ A negative exponent moves the power to the other side of the fraction bar, for a nonzero base:

- 7 e^{- 7 y}

- \frac{7}{e^{7 y}}

⇥

- \frac{7}{e^{7 y}}

⇥1.7) Then you can continue the derivation:

⇥ⅰ

\frac{d}{d y} (\frac{1}{e^{7 y}})

- \frac{7}{e^{7 y}}

⇥

- \frac{7}{e^{7 y}}

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥ⓘ The derivative of

- \frac{1}{7} \cdot \frac{1}{e^{7 y}}

\frac{1}{e^{7 y}}

, so

- \frac{1}{7} \cdot \frac{1}{e^{7 y}}

is an antiderivative of

\frac{1}{e^{7 y}}

\int \frac{1}{e^{7 y}} d y

- \frac{1}{7} \cdot \frac{1}{e^{7 y}}

⇥

- \frac{1}{7} \cdot \frac{1}{e^{7 y}}

⇥ⓘ To multiply fractions, multiply the numerators together and the denominators together:

- \frac{1}{7} \cdot \frac{1}{e^{7 y}}

- \frac{1 \cdot 1}{7 e^{7 y}}

- \frac{1}{7 e^{7 y}}

⇥

- \frac{1}{7 e^{7 y}}

⇥1.8) Let's compute the integral:

\int \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

⇥For the math expression:

\frac{e^{\sqrt{x}}}{\sqrt{x}}

, since (exp(a))' = a'*exp(a), we try differentiating

\exp (\sqrt{x})

⇥1.9) Let's perform the differentiation:

\frac{d}{d x} (\exp (\sqrt{x}))

⇥Apply the differentiation rule:

\frac{d}{d x} (\exp (f)) = \frac{d}{d x} (f) \cdot \exp (f)

with f =

\sqrt{x}

⇥ⓘ The derivative of exp(x) is exp(x):

\frac{d}{d x} (\exp (\sqrt{x}))

\frac{d}{d x} (\sqrt{x}) \cdot \exp (\sqrt{x})

⇥

\frac{d}{d x} (\sqrt{x}) \cdot \exp (\sqrt{x})

⇥ⓘ The derivative of sqrt(x) is 1/(2*sqrt(x)):

\frac{d}{d x} (\sqrt{x})

\frac{1}{2 \sqrt{x}}

⇥

\frac{1}{2 \sqrt{x}} \cdot \exp (\sqrt{x})

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2 \sqrt{x}} \cdot \exp (\sqrt{x})

\frac{1 \cdot \exp (\sqrt{x})}{2 \sqrt{x}}

\frac{\exp (\sqrt{x})}{2 \sqrt{x}}

⇥

\frac{\exp (\sqrt{x})}{2 \sqrt{x}}

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥ⓘ The derivative of

2 e^{\sqrt{x}}

\frac{e^{\sqrt{x}}}{\sqrt{x}}

, so

2 e^{\sqrt{x}}

is an antiderivative of

\frac{e^{\sqrt{x}}}{\sqrt{x}}

\int \frac{e^{\sqrt{x}}}{\sqrt{x}} d x

2 e^{\sqrt{x}}

⇥

2 e^{\sqrt{x}}

Integrating both sides gives

- \frac{1}{7 e^{7 y}} = 2 e^{\sqrt{x}} + C

, where

C

is an arbitrary real constant

Multiply by

-7

and rename

- 7 C

C

e^{-7 y} = - 14 e^{\sqrt{x}} + C

Isolate

y

-7 y = \ln (- 14 e^{\sqrt{x}} + C)

, hence

y = - \frac{\ln (- 14 e^{\sqrt{x}} + C)}{7}

, where

- 14 e^{\sqrt{x}} + C > 0

▶Solution:

y = - \frac{\ln (- 14 e^{\sqrt{x}} + C)}{7}