Question

Y'+y=y^2; y(0)=-1/3 Giải phương trình trên

Answer 1

Step-by-step explanation:

$y' + y = y^2$

We can rewrite the differential equation above as

$\dfrac{dy}{dx} + y = y^2$

$dy = (y^2 - y)dx$

or

$\dfrac{dy}{y^2 -y} = dx$

We can rewrite the left side of the equation above as

$\dfrac{dy}{y^2-y}=\dfrac{dy}{y(y-1)}= \left(\dfrac{1}{y-1} - \dfrac{1}{y} \right)dy$

We can the easily integrate this as

$\displaystyle \int \left(\dfrac{1}{y-1} - \dfrac{1}{y} \right)dy = \int dx$

or

$\displaystyle \int \dfrac{dy}{y-1} - \int \dfrac{dy}{y} = \int dx$

This will then give us

$\ln |y-1| - \ln |y| + \ln |k| = x$

where k is the constant of integration. Combining the terms on the left hand side, we get

$\ln \left|\dfrac{k(y-1)}{y} \right| = x$

or

$\dfrac{y-1}{y} = \frac{1}{k}e^x$

Solving for y, we get

$y= \dfrac{1}{1- \frac{1}{k} e^x}=\dfrac{k}{k-e^x}$

We know that $y(0)= \frac{1}{3}$, so when we substitute $x=0$, we find that $k = -\frac{1}{2}$.

Therefore, the final form of the solution to the differential equation above is

$y = \dfrac{1}{1+2e^x}$