What is the general solution of the differential equation $(1 + x) \frac{d y}{d x} - y = 1 + x$ $(1+x)dy/dx-y=1+x$ ?

Question

1475 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T16:14:52

$y = (1 + x) ln (1 + x) + C (1 + x)$ $y = (1+x)ln(1+x) + C(1+x)$

We have:

$(1 + x) \frac{d y}{d x} - y = 1 + x$ $(1+x)dy/dx-y=1+x$

We can re-arrange this ODE as follows:

$\frac{d y}{d x} - \frac{1}{1 + x} y = 1$ $dy/dx - 1/(1+x) y=1$ ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{d y}{d x} + P (x) y = Q (x)$ $dy/dx + P(x)y=Q(x)$

We can readily generate an integrating factor when we have an equation of this form, given by;

$I = e^{\int P (x) d x}$ $I = e^(int P(x) dx)$
$\ \ = exp(int \ -1/(1+x) \ dx)$
$\ \ = exp( -ln(1+x) )$
$\ \ = -(1+x)$

And if we multiply the DE [1] by this Integrating Factor, $I$ , we will have a perfect product differential;

$-1/(1+x) dy/dx + 1/(1+x)^2 y = -1/(1+x)$
$:. d/dx[ -1/(1+x) y] = -1/(1+x)$

Which we can directly integrate to get:

$1/(1+x) y = int \ 1/(1+x) \ dx$
$:. 1/(1+x) y = ln(1+x) + C$

$:. y = (1+x)ln(1+x) + C(1+x)$

What is the general solution of the differential equation (1+x)dy/dx-y=1+x (1+x)dydx−y=1+x (1+x)dy/dx-y=1+x ?