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,