Question

Find the solution of \[ydx - xdy + \log xdx = 0\]
A. \[y - \log x - 1 = C\]
B. \[x + \log y + 1 = Cx\]
C. \[y + \log x + 1 = Cx\]
D. \[y + \log x - 1 = Cx\]

Answer 1

Hint: First we rewrite the given equation with the form of \[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\] . Then we compare the given differential equation with the general equation to find the \[P(x)\] and apply the IF formula to calculate the integrating factor of the given differential equation.

Formula Used:
The general form of Linear Differential Equation \[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\], \[y.IF = \int {Q.IFdx} + C\]and Integration by parts is \[\int {u{\rm{ }}v{\rm{ }}dx} {\rm{ }} = {\rm{ }}u\int v {\rm{ }}dx{\rm{ }} - {\rm{ }}\int {\dfrac{{du}}{{dx}}{\rm{ }}\left( {\int v {\rm{ }}dx} \right)} {\rm{ }}dx\].

Complete step by step solution:
Separating \[dx\] and \[dy\] in RHS and LHS we will get,
\[ydx - xdy + \log xdx = 0\]
\[ \Rightarrow xdy = \left( {y + \log x} \right)dx\]
Converting the equation in the form of \[\dfrac{{dy}}{{dx}}\] we will get,
\[\dfrac{{dy}}{{dx}} = \dfrac{y}{x} + \dfrac{{\log x}}{x}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{x} = \dfrac{{\log x}}{x}\] ----- (i)
From equation (i) we can clearly see that it is the Linear Differential Equation in terms of \[y\] where the general form of Linear Differential Equation is \[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\] where \[P(x) = \left( { - \dfrac{1}{x}} \right)\] and \[Q(x) = \dfrac{{\log x}}{x}\] .
Now we will be finding the Integrating Factor whose formula is \[{e^{\int {Pdx} }}\] .
\[{e^{\int {Pdx} }}\]
\[ = {e^{\int {\dfrac{{ - 1}}{x}dx} }}\]
Integrating and we get
\[ = {e^{ - \log x}}\]
Using the Property of \[\log \] and we get
\[ = {e^{\log {x^{ - 1}}}}\]
Therefore \[{e^{\log {x^{ - 1}}}}\] is equal to \[{x^{ - 1}}\]
Integrating factor,
 \[IF = \dfrac{1}{x}\]
According to linear differential theorem,
\[y \cdot IF = \int {Q(x) \cdot IFdx} + C\] , where \[C\] is a constant of integration.
Put the value of \[IF\] and \[Q(x)\] and integrate the following,
\[ \Rightarrow y \cdot \dfrac{1}{x} = \int {\dfrac{{\log x}}{x} \times \dfrac{1}{x}dx} + C\]
\[ \Rightarrow \dfrac{y}{x} = \int {\log x \times \dfrac{1}{{{x^2}}}dx} + C\] ………………..(ii)
Now, to integrate the equation (ii) we will be using Integration by Parts to proceed further,
Integration By Part - \[\int {u{\rm{ }}v{\rm{ }}dx} {\rm{ }} = {\rm{ }}u\int v {\rm{ }}dx{\rm{ }} - {\rm{ }}\int {\dfrac{{du}}{{dx}}{\rm{ }}\left( {\int v {\rm{ }}dx} \right)} {\rm{ }}dx\]
Where, \[u = \log x\] and \[v = \dfrac{1}{{{x^2}}}\]
Now, substituting the values of \[u\] and \[v\] in the formula, we will get,
\[\int {u{\rm{ }}v{\rm{ }}dx} {\rm{ }} = {\rm{ }}u\int v {\rm{ }}dx{\rm{ }} - {\rm{ }}\int {\dfrac{{du}}{{dx}}{\rm{ }}\left( {\int v {\rm{ }}dx} \right)} {\rm{ }}dx\]
\[ \Rightarrow \int {logx \times \dfrac{1}{{{x^2}}}{\rm{ }}dx} {\rm{ }} = {\rm{ lo}}gx\int {\dfrac{1}{{{x^2}}}{\rm{ }}dx} {\rm{ }} - {\rm{ }}\int {\dfrac{{d\left( {logx} \right)}}{{dx}}{\rm{ }}\left( {\int {\dfrac{1}{{{x^2}}}{\rm{ }}dx} } \right)} {\rm{ }}dx\]
After differentiating and integrating the terms in the above equation we will get,
\[ \Rightarrow \dfrac{y}{x} = logx \times \left( { - \dfrac{1}{x}} \right) - \int {\dfrac{1}{x}} \times \left( { - \dfrac{1}{x}} \right)dx + C\]
\[ \Rightarrow \dfrac{y}{x} = - \dfrac{{logx}}{x} + \left( { - \dfrac{1}{x}} \right) + C\]
\[ \Rightarrow \dfrac{y}{x} = \dfrac{{ - logx - 1 + Cx}}{x}\]
\[ \Rightarrow y = - logx - 1 + cx\]
\[ \Rightarrow y + \log x + 1 = Cx\]
Hence the correct option is (C).

Note: Students tries to the given equation directly without using the format \[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\] and for this reason they unable to solve the given differential equation.