Question

Solve the initial value problem: dydx+5y=7 y(0)=0

Answer 1

Answer:

Given differential equation,

$\frac{dy}{dx}+5y=7$

$\frac{dy}{dx}=7-5y$

$\implies \frac{dy}{7-5y}=dx$

Taking integration both sides,

$\int \frac{dy}{7-5y}=\int dx$

Put 7 - 5y = u ⇒ -5 dy = du ⇒ dy = -du/5,

$-\frac{1}{5} \int \frac{du}{u} = \log x + C$

$-\frac{1}{5} \log u = \log x + C$

$-\frac{1}{5}\log(7-5y) = \log x + C---(1)$

Here, x = 0, y = 0

$\implies -\frac{1}{5} \log 7= C$

Hence, from equation (1),

$-\frac{1}{5}\log(7-5y)=\log x -\frac{1}{5}log 7$

$\log(7-5y)=\log (\frac{x}{7^\frac{1}{5}})$

$7-5y=\frac{x}{7^\frac{1}{5}}$

$7-\frac{x}{7^\frac{1}{5}}=5y$

$\implies y=\frac{1}{5}(7-\frac{x}{7^\frac{1}{5}})$