# Thread: Solving dy/dx + (y/x) = y^3

1. ## Solving dy/dx + (y/x) = y^3

Hi,

Help needed. How is (a) done?

(a) The variables x and y are related by the differential equation $\displaystyle \frac{dy}{dx}\,+\,\frac{y}{x}\,=\,y^3$

(i) State clearly why the integrating factor method cannot be used to solve this equation.

(ii) The variables y and z are related by the equation $\displaystyle \frac{1}{y^2}\,=\,-2z$. Show that $\displaystyle \frac{dz}{dx}\,-\,\frac{2z}{x}\,=\,1$.

(iii) Find the solution of the differential equation $\displaystyle \frac{dy}{dx}\,+\,\frac{y}{x}\,=\,y^3,\,$ given that $\displaystyle y\,=\,2\,\,when\,\,x\,=\,1$.

(b) The rate at which atoms in a mass of radioactive material are disintegrating is proportional to n, the number of atoms present at any time time t, measured in days. Initially the number of atoms is m.

(i) Form and solve the differential equation which represents this data.

$\displaystyle \frac{dn}{dt} = -kn$

$\displaystyle \int \frac{dn}{n} = \int -k dt$

$\displaystyle \ln(n) = -kt + C$

$\displaystyle = Ae^{-kt}$

$\displaystyle t = 0$
$\displaystyle n = m$

$\displaystyle m = Ae^0$

$\displaystyle \therefore n = me^{-kt}$

(ii) Given that half of the original mass disintegrates in 76 days, evaluate the constant of proportion in the differential equation.

t = 76 days

$\displaystyle n = \frac{1}{2}m$

$\displaystyle \frac{1}{2}m = me^{-76k}$

$\displaystyle \frac{1}{2m}m = e^{-76k}$

$\displaystyle \ln \frac{1}{2m}m = -76k$

$\displaystyle \Rightarrow \frac{1}{76} \ln 2 = k$

2. Originally Posted by Hellbent
Hi,

Help needed. How is (a) done?

(a) The variables x and y are related by the differential equation $\displaystyle \frac{dy}{dx}\,+\,\frac{y}{x}\,=\,y^3$$\displaystyle$

(i) State clearly why the integrating factor method cannot be used to solve this equation. Mr F says: What is the general form of DE for which the integrating factor method can be used ...? Compare this to what you have.

(ii) The variables y and z are related by the equation $\displaystyle \frac{1}{y^2}\,=\,-2z$. Show that $\displaystyle \frac{dz}{dx}\,-\,\frac{2z}{x}\,=\,1$. Mr F says: Substitute -2z = 1/y^2 into the given DE.

(iii) Find the solution of the differential equation $\displaystyle \frac{dy}{dx}\,+\,\frac{y}{x}\,=\,y^3,\,$ given that $\displaystyle y\,=\,2\,\,when\,\,x\,=\,1$. Mr F says: Solve the DE in part (ii) and then back-substitute -2z = 1/y^2.
[snip]
If you need more help, please say where you are stuck.