Question

The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^xdx is ______.

Options

• (y + 1) = k(e^x + 1)

• y + 1 = e^x + 1 + k

• y = log {k(y + 1)(e^x + 1)}

• y = ${\displaystyle \log \left\{\frac{{\text{e}}^x+1}{y+1}\right\}+\text{k}}$

Answer 1

The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^xdx is y = log {k(y + 1)(e^x + 1)}.

Explanation:

The given differential equation is (e^x + 1) ydy = (y + 1) e^xdx

⇒ ${\displaystyle \frac{y}{y+1}\text{d}y=\frac{{\text{e}}^x}{{\text{e}}^x+1}\text{d}x}$

Integrating both sides, we get

${\displaystyle \int \frac{y}{y+1}\text{d}y=\int \frac{{\text{e}}^x}{{\text{e}}^x+1}\text{d}x}$

⇒ ${\displaystyle \int \frac{y+1-1}{y+1}\text{d}y=\int \frac{{\text{e}}^x}{{\text{e}}^x+1}\text{d}x}$ 

⇒ ${\displaystyle \int 1.\text{d}y-\int \frac{1}{y+1}\text{d}y=\int \frac{{\text{e}}^x}{{\text{e}}^x+1}\text{d}x}$

⇒ ${\displaystyle y-\log |y+1|=\log |{\text{e}}^x+1|+\log \text{k}}$

⇒ y = ${\displaystyle \log |y+1|+\log |{\text{e}}^x+1|+\log \text{k}}$

⇒ y = ${\displaystyle \log \left|\text{k}(y+1)({\text{e}}^x+1)\right|}$