Question

How to solve this first order linear differential equation?
 $
  xy' - \dfrac{1}{{x + 1}}y = x \\
  y\left( 1 \right) = 0 \;
  $

Answer 1

Hint: We have to solve the given first order differential equation. We can observe that the differential equation can be written in the form of $ y' + P\left( x \right)y = Q\left( x \right) $ . To solve such differential equations we use an integrating factor which is given as $ I.F. = {e^{\int {P\left( x \right)dx} }} $ . Here $ P\left( x \right) $ is the same function which is multiple of $ y $ is the given differential equation. We multiply both sides of the differential equation by the integrating factor and further integrate to find the solution. We can then use the given condition \[y\left( 1 \right) = 0\] to find the value of the integrating constant.

Complete step-by-step answer:
We have been given to solve the first order differential equation $ xy' - \dfrac{1}{{x + 1}}y = x $ with the condition \[y\left( 1 \right) = 0\].
The given differential equation can be written as,
 $
  xy' - \dfrac{1}{{x + 1}}y = x \\
   \Rightarrow y' + \left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right)y = 1 \;
  $
This differential equation is now in the form of $ y' + P\left( x \right)y = Q\left( x \right) $ where $ P\left( x \right) = \left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right) $ and $ Q\left( x \right) = 1 $ .
To solve such differential equations we use an integrating factor given as $ I.F. = {e^{\int {P\left( x \right)dx} }} $ .
We can find the $ I.F. $ as,
 $ {e^{\int {P\left( x \right)dx} }} = {e^{\int {\left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right)dx} }} $
By partial fractions we can write,
 $
  \left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right) = \dfrac{a}{x} + \dfrac{b}{{x + 1}} \\
   \Rightarrow - 1 = a\left( {x + 1} \right) + bx \;
  $
If $ x = 1 $ , then $ 2a + b = - 1 $
If $ x = 2 $ , then $ 3a + 2b = - 1 $
On solving we get $ a = - 1 $ and $ b = 1 $ .
Thus,
 $ {e^{\int {\left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right)dx} }} = {e^{\int {\left( { - \dfrac{1}{x} + \dfrac{1}{{\left( {x + 1} \right)}}} \right)dx} }} = {e^{ - \ln x + \ln \left( {x + 1} \right)}} = {e^{\ln \left( {1 + \dfrac{1}{x}} \right)}} = \left( {1 + \dfrac{1}{x}} \right) $
Thus, $ I.F. = \left( {1 + \dfrac{1}{x}} \right) $
We multiply the integrating factor on both sides of the equation and integrate.
 $ \Rightarrow \left( {1 + \dfrac{1}{x}} \right)y' + \left( {1 + \dfrac{1}{x}} \right)\left( { - \dfrac{1}{{x\left( {x + 1} \right)}}} \right)y = \left( {1 + \dfrac{1}{x}} \right) $
This can be written as,
 $ \Rightarrow \dfrac{{d\left\{ {\left( {1 + \dfrac{1}{x}} \right)y} \right\}}}{{dx}} = \left( {1 + \dfrac{1}{x}} \right) $
Now we integrate both sides.
 $
   \Rightarrow \int {\dfrac{{d\left\{ {\left( {1 + \dfrac{1}{x}} \right)y} \right\}}}{{dx}}dx} = \int {\left( {1 + \dfrac{1}{x}} \right)dx} \\
   \Rightarrow \left( {1 + \dfrac{1}{x}} \right)y = x + \ln x + C \;
  $
We can find the value of $ C $ using the condition \[y\left( 1 \right) = 0\].
 $
   \Rightarrow \left( {1 + 1} \right) \times 0 = 1 + \ln 1 + C \\
   \Rightarrow C + 1 = 0 \\
   \Rightarrow C = - 1 \;
  $
Thus we get,
 $ \left( {1 + \dfrac{1}{x}} \right)y = x + \ln x - 1 $
Or we can write the function as,
 $ y = \dfrac{{x\left( {x + \ln x - 1} \right)}}{{\left( {x + 1} \right)}} $
So, the correct answer is “ $ y = \dfrac{{x\left( {x + \ln x - 1} \right)}}{{\left( {x + 1} \right)}} $ ”.

Note: We solved the differential equation using the integrating factor. The integrating factor of the form $ I.F. = {e^{\int {P\left( x \right)dx} }} $ is used for the differential equation of the form $ y' + P\left( x \right)y = Q\left( x \right) $ and may not work properly for other forms of differential equation. We have to be careful while following each step in the solution.