(dy)/(dx)+3y=40 — step-by-step solution

⚠ (dy)/(dx)+3y=40 is read as y'+3y=40: use dif(expression, variable) for a derivative with respect to that variable, use (numerator)/(denominator) for a quotient

Solve the linear equation

y' + 3 y = 40

, where

y

is a function of

x

. The polynomial coefficients are continuous on ℝ

The form

y' + p y = q

gives

p = 3

and

q = 40

⇥1) Let's compute the integral:

\int 3 d x

⇥ⓘ The derivative of

3 x

3

, so

3 x

is an antiderivative of

3

\int 3 d x

3 x

⇥

3 x

The integrating factor exp(∫p) is

\exp (3 x)

, strictly positive on ℝ

Multiplying by this factor, the product rule gives

{(\exp (3 x) y)}^{'}

40 \exp (3 x)

⇥2) Let's integrate:

\int 40 \exp (3 x) d x

⇥For the math expression:

40 \exp (3 x)

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

\exp (3 x)

⇥3) Let's perform the differentiation:

\frac{d}{d x} (\exp (3 x))

⇥Apply the differentiation rule:

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

with f =

3 x

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

\frac{d}{d x} (\exp (3 x))

\frac{d}{d x} (3 x) \cdot \exp (3 x)

⇥

\frac{d}{d x} (3 x) \cdot \exp (3 x)

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

\frac{d}{d x} (3 x)

3

⇥

3 \exp (3 x)

⇥The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

⇥ⓘ The derivative of

\frac{40}{3} \cdot \exp (3 x)

40 \exp (3 x)

, so

\frac{40}{3} \cdot \exp (3 x)

is an antiderivative of

40 \exp (3 x)

\int 40 \exp (3 x) d x

\frac{40}{3} \cdot \exp (3 x)

⇥

\frac{40}{3} \cdot \exp (3 x)

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

\frac{40}{3} \cdot \exp (3 x)

\frac{40 \exp (3 x)}{3}

⇥

\frac{40 \exp (3 x)}{3}

Integrating gives

\exp (3 x) y = \frac{40 \exp (3 x)}{3} + C

. Divide by the factor, which is never zero;

C

is an arbitrary real constant, including zero

▶Solution:

y = \frac{40}{3} + \frac{C}{\exp (3 x)}