Question

Find the number of solutions of the differential equation \[y' = \dfrac{{y + 1}}{{x - 1}}\], if \[y\left( 1 \right) = 2\].
A. None
B. One
C. Two
D. Infinite

Answer 1

Hint: Here, a first-order differential equation is given. First, rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\] and simplify the differential equation. Then, integrate both sides with respect to the corresponding variables. After that, solve the integrals and find the solution of the differential equation. Then, substitute the values from the given equation \[y\left( 1 \right) = 2\] in the solution of the differential equation and find the value of the integration constant \[c\]. In the end, verify the conditions and find the required answer.

Formula Used: \[\int {\dfrac{1}{x}} dx = \log x + c\]
\[\log a + \log b = \log \left( {ab} \right)\]

Complete step by step solution: Given:
The first-order differential equation is \[y' = \dfrac{{y + 1}}{{x - 1}}\] and \[y\left( 1 \right) = 2\].

Let’s solve the differential equation.
Rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{y + 1}}{{x - 1}}\]
Simplify the equation.
\[\dfrac{{dy}}{{y + 1}} = \dfrac{{dx}}{{x - 1}}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{dy}}{{y + 1}}} = \int {\dfrac{{dx}}{{x - 1}}} \]
Apply the integration formulas \[\int {\dfrac{1}{x}} dx = \log x + c\].
\[\log \left( {y + 1} \right) = \log \left( {x - 1} \right) + \log c\]
Apply the product property of the logarithm on the right-hand side.
\[\log \left( {y + 1} \right) = \log \left[ {\left( {x - 1} \right)c} \right]\]
Equate both sides.
\[\left( {y + 1} \right) = \left( {x - 1} \right)c\]
\[ \Rightarrow \dfrac{{y + 1}}{{x - 1}} = c\] \[.....\left( 1 \right)\]

Now to calculate the value of the integration constant, substitute the values of the given equation \[y\left( 1 \right) = 2\] in the equation \[\left( 1 \right)\].
We have, \[x = 1\] and \[y = 2\]
Then,
\[\dfrac{{2 + 1}}{{1 - 1}} = c\]
\[ \Rightarrow \dfrac{3}{0} = c\]
\[ \Rightarrow c = 0\]

So, the required solution is possible only when \[x - 1 = 0\].
Therefore, only one solution exists for the given differential equation \[y' = \dfrac{{y + 1}}{{x - 1}}\].

Option ‘B’ is correct

Note: Students often apply a wrong formula to integrate \[\dfrac{1}{x}\] . They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].