Find y' for e^(y/x)ln x - c e^(y/x) + 1 = 0, where c is

Find $y'$ for $e^{y/x}\ln x - c\,e^{y/x} + 1 = 0$, where $c$ is a constant.

Step 1: Factor the relation

Both exponential terms share the factor $e^{y/x}$: \[e^{y/x}\bigl(\ln x - c\bigr) + 1 = 0 \;\Longrightarrow\; e^{y/x}\bigl(\ln x - c\bigr) = -1. \tag{1}\]

Keeping this form in reserve is what makes the computation short.

Step 2: Differentiate with respect to $x$

Write $u = \dfrac{y}{x}$, so that $u' = \dfrac{y'x-y}{x^{2}}$ and $\bigl(e^{u}\bigr)' = e^{u}u'$. Differentiating the two exponential terms together, and the $\ln x$ separately:

\[ e^{y/x}u'\bigl(\ln x - c\bigr) + e^{y/x}\cdot\frac1x = 0. \]

Step 3: Substitute the relation (1)

The bracket combination is already known to equal $-1$: $(-1)\cdot u' + \frac{e^{y/x}}{x} = 0 \;\Longrightarrow\; u' = \frac{e^{y/x}}{x}$.

Step 4: Unpack $u'$

\[ \frac{y'x-y}{x^{2}} = \frac{e^{y/x}}{x} \;\Longrightarrow\; y'x - y = x\,e^{y/x} \;\Longrightarrow\; y' = \frac{y}{x} + e^{y/x}. \]

Step 5: Verification

From (1), $e^{y/x} = \dfrac{1}{c-\ln x}$, so the relation can be solved explicitly: $\frac{y}{x} = -\ln\bigl(c-\ln x\bigr) \;\Longrightarrow\; y = -x\ln\bigl(c-\ln x\bigr)$.

\[y' = -\ln\bigl(c-\ln x\bigr) + \frac{1}{c-\ln x} = \frac{y}{x} + e^{y/x} \;\;\checkmark\]

\[c = 1,\ x = 1:\quad y = 0,\ e^{y/x} = 1 \;\Longrightarrow\; y' = 0+1 = 1 \;\;\checkmark\]