Question

How to solve this first order linear differential equation?

`xy'-1/(x+1)y=x`
y(1) = 0

(According to our professor, I.F. = `e^(intf(x))`, and we should just leave the integral be for now if the integral cannot be solved by hand or conventional methods)

Answer 1

`y = x/(x+1)(x + lnx -1)`

Explanation:

We have:

`xy'-1/(x+1)y=x` with `y(1) = 0`

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

So, we can put the equation in standard form:

`y'-1/(x(x+1))y = 1`

Then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -1/(x(x+1)) \ dx)`

We can readly evaluate this integral if we perform a partial fraction decomposition of the integrand:

`1/(x(x+1)) -= A/x + B/(x+1) => 1 -= A(x+1)+Bx`

Then:

`x= \ \ \ \ \ 0 => A=1`
`x=-1 => B=-1`

So we can write:

`I = exp(int \ -1/x + 1/(x+1) ) \ dx)`
`\ \ = exp(ln(x+1)-lnx)`
`\ \ = exp( ln((x+1)/x) )`
`\ \ = (x+1)/x`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will have a perfect product differential (in fact the original equation);

`:. ((x+1)/x)y' - ((x+1)/x) 1/(x(x+1))y = ((x+1)/x)`

`:. (1+1/x)y' - 1/x^2y = 1+1/x`

`:. d/dx ((1+1/x)y) = 1+1/x`

This is now separable, so by "separating the variables" we get:

`(1+1/x)y = int \ 1+1/x \ dx`

Which is trivial to integrate to get the General Solution:

`(1+1/x)y = x + lnx + C`

Applying the initial condition `y(1)=0` we get:

`0 = 1 +ln1 + C => C=-1`

Leading to the Particular Solution:

`(x+1)/x \ y = x + lnx -1`

`:. y = x/(x+1)(x + lnx -1)`

Answer 2

`y=x/(x+1)(x+lnx-1)`

Explanation:

.

`xy'-1/(x+1)y=x`

A first Order linear Differential Equation has the form of:

`y'(x)+p(x)y=q(x)`

Let's put our ODE in this form by dividing the entire equation by `x`:

`y'-1/(x(x+1))y=1`, which shows that:

`p(x)=-1/(x(x+1))` and `q(x)=1`

The integration factor is:

`mu(x)=e^(intp(x)dx)=e^(int-1/(x(x+1))dx)=e^I`

`I=int-1/(x(x+1))dx=-int1/(x(x+1))dx`

We can use partial fraction expansion to solve it:

`1/(x(x+1))=A/x+B/(x+1)=(A(x+1)+Bx)/(x(x+1))=(Ax+A+Bx)/(x(x+1))=((A+B)x+A)/(x(x+1))`

`A+B=0` and `A=1`

Therefore, `B=-1`

`I=-int(1/x-1/(x+1))dx=-lnabsx+lnabs(x+1)+C`

`mu(x)=e^(-lnabsx+ln(x+1)+C)=e^-lnabsx*e^(lnabs(x+1))*e^C=1/e^lnabsx(e^lnabs(x+1))e^C=1/x(x+1)e^C=(e^C(x+1))/x`

The constant `e^C` is redundant and can be ignored. Therefore,our integration factor is:

`mu(x)=(x+1)/x`

Now, we multiply both sides of our ODE by this integration factor:

`y'((x+1)/x)-1/(x(x+1))((x+1)/x)y=(x+1)/x`

Then, we simplify and refine:

`((x+1)y')/x-y/x^2=1/x+1` `color(red)(Equation-1)`

If `f=(x+1)/x and g=y` then applying the product rule of differentiation we get:

`(f*g)'=f*g'+f'*g`

`f'=(x-(x+1))/x^2=-1/x^2` (we used the quotient rule)

`g'=y'`

`f*g'+f'*g=((x+1)/x)y'+(-1/x^2)y=((x+1)y')/x-y/x^2`

which is the same as Left Hand Side of `color(red)(Equation-1)`.

Therefore, the Right Hand Side of it must be equal to `(f*g)'`.

`(f*g)'=1/x+1`

`(((x+1)y)/x)'=1/x+1`, i.e:

`d/dx((x+1)/xy)=1/x+1`

We now take the integral of both sides:

`(x+1)/xy=int(1/x+1)dx=int1/xdx+intdx`

`(x+1)/xy=lnx+x+c`

We now proceed to isolate `y`:

Let's multiply both sides by `x`:

`(x+1)y=xlnx+x^2+cx`

Let's divide both sides by `x+1`:

`y=(xlnx+x^2+cx)/(x+1)`

Now, we can apply the initial conditions:

`y(1)=0`

`((1)(ln1)+(1)^2+c(1))/((1)+1)=((1)(0)+1+c)/2=(c+1)/2=0`

`c+1=0`

`c=-1`

Therefore,

`y=(xlnx+x^2-x)/(x+1)=x/(x+1)(x+lnx-1)`