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.