# 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)

y(1) = 0

(According to our professor, I.F. =

##### 2 Answers

# 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,

# :. ((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

# 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) #

#### Explanation:

.

A first Order linear Differential Equation has the form of:

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

The integration factor is:

We can use partial fraction expansion to solve it:

Therefore,

The constant

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

Then, we simplify and refine:

If

which is the same as Left Hand Side of

Therefore, the Right Hand Side of it must be equal to

We now take the integral of both sides:

We now proceed to isolate

Let's multiply both sides by

Let's divide both sides by

Now, we can apply the initial conditions:

Therefore,