Question

The solution of \[x\dfrac{{dy}}{{dx}} = y + x{e^{\dfrac{y}{x}}}\] with \[y(1) = 0\] is
A. \[{e^{\dfrac{y}{x}}} + \ln x = 1\]
B. \[{e^{\dfrac{{ - y}}{x}}} = \ln x\]
C. \[{e^{\dfrac{{ - y}}{x}}} + 2\ln x = 1\]
D. \[{e^{\dfrac{{ - y}}{x}}} + \ln x = 1\]

Answer 1

Hint: We take all the values to one side of the equation and keep \[\dfrac{{dy}}{{dx}}\] on other side of the equation, then substitute the power of e as a new variable and write the equation in terms of that new variable. Using the separation method, write the terms in each variable along with their respective derivatives and integrate for the solution. Finally substitute the given value in the final solution and choose the correct option.

Complete step-by-step answer:
We have the differential equation as \[x\dfrac{{dy}}{{dx}} = y + x{e^{\dfrac{y}{x}}}\].
Divide both sides of the equation by \[x\].
\[
   \Rightarrow \dfrac{x}{x} \times \dfrac{{dy}}{{dx}} = \dfrac{{y + x{e^{\dfrac{y}{x}}}}}{x} \\
   \Rightarrow \dfrac{x}{x} \times \dfrac{{dy}}{{dx}} = \dfrac{y}{x} + \dfrac{{x{e^{\dfrac{y}{x}}}}}{x} \\
 \]
Cancel out the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} + {e^{\dfrac{y}{x}}}\] … (1)
Now we assume the power of e as a new variable.
Let \[\dfrac{y}{x} = t\]
Then cross multiplying the denominator of LHS to RHS we get,
\[y = tx\]
Differentiating both sides of the equation with respect to x we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(tx)\]
Since t is also a function of x, we use product rule to differentiate RHS of the equation.
Product rule of differentiation gives us\[\dfrac{d}{{dx}}(mn) = m\dfrac{{dn}}{{dx}} + n\dfrac{{dm}}{{dx}}\] .
\[ \Rightarrow \dfrac{{dy}}{{dx}} = t \times \dfrac{{dx}}{{dx}} + x\dfrac{{dt}}{{dx}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = t + x\dfrac{{dt}}{{dx}}\] … (2)
Now we substitute the value of \[\dfrac{y}{x} = t\] and \[\dfrac{{dy}}{{dx}} = t + x\dfrac{{dt}}{{dx}}\]from equation (2) in equation (1)
\[ \Rightarrow t + x\dfrac{{dt}}{{dx}} = t + {e^t}\]
Shift the value of t from LHS to RHS of the equation.
\[
   \Rightarrow x\dfrac{{dt}}{{dx}} = t - t + {e^t} \\
   \Rightarrow x\dfrac{{dt}}{{dx}} = {e^t} \\
 \]
Now using separation of variables, we write the variables and their derivatives on separate sides.
Cross multiply both sides of the equation.
 \[ \Rightarrow \dfrac{{dt}}{{{e^t}}} = \dfrac{{dx}}{x}\]
Now we can write \[\dfrac{1}{{{e^t}}} = {e^{ - t}}\]
\[ \Rightarrow {e^{ - t}}dt = \dfrac{{dx}}{x}\]
Now we integrate both sides of the equation.
\[ \Rightarrow \int {{e^{ - t}}dt} = \int {\dfrac{1}{x}dx} \]
Shift all the integration to one side of the equation
\[ \Rightarrow \int {{e^{ - t}}dt} - \int {\dfrac{1}{x}dx} = 0\] … (3)
Since, we know that \[\int {{e^{ - t}}dt = - {e^{ - t}}} \] and \[\int {\dfrac{1}{x}dx} = \log x\]
Substitute the values in equation (3)
\[ \Rightarrow - {e^{ - t}} - \ln x + C = 0\] {where C is constant of integration}
Substitute the value of \[t = \dfrac{y}{x}\] back in the equation.
\[ \Rightarrow - {e^{\dfrac{{ - y}}{x}}} - \ln x + C = 0\]
So to find the value of C we shift all the values except C to one side of the equation.
\[ \Rightarrow {e^{\dfrac{{ - y}}{x}}} + \ln x = C\] … (4)
This is the general solution of the differential equation.
Now we calculate the particular solution.
We are given that \[y(1) = 0\] which means that when we substitute the value of \[x = 1\] then we get \[y = 0\].
Substitute the values of x and y in equation (4)
\[
   \Rightarrow {e^{\dfrac{{ - 0}}{1}}} + \ln 1 = C \\
   \Rightarrow {e^0} + 0 = C \\
   \Rightarrow 1 = C \\
 \] {Substituting the value of \[\ln 1 = 0\] }
Substituting the value of C=1 in equation (4)
\[ \Rightarrow {e^{\dfrac{{ - y}}{x}}} + \ln x = 1\]

So, the correct answer is “Option D”.

Note: Students many times make mistakes while differentiating the new variable that is assumed, they don’t differentiate the variable with respect to x which is wrong. Since we have a product of x and a new variable t we have to apply the product rule there because the new variable is also a function of x.