Question

The general solution of the differential equation \( x\,dy + \left(y - e^x\right) dx = 0 \) is:

• A. \( e^{xy} + e^x = C \), Where \(C\) is constant of integration
• B. \( \frac{x^2}{2} + xy - e^x = C \), Where \(C\) is constant of integration
• C. \( \frac{x^2}{2} + \frac{y^2}{2} - e^x = C \), Where \(C\) is constant of integration
• D. \( xy - e^x = C \), Where \(C\) is constant of integration

Hint:

For exact differential equations \(M\,dx + N\,dy = 0\), check if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). If yes, find \(\Psi\) by integrating \(M\) with respect to \(x\) and use \(N\) to determine the remaining function of \(y\).

Answer 1

The Correct Option is B

Given the differential equation: \[ x\,dy + \left(y - e^x\right) dx = 0 \] Rewrite it as: \[ \left(y - e^x\right) dx + x\,dy = 0 \] Let \(M = y - e^x\) and \(N = x\). The equation is \( M\,dx + N\,dy = 0 \). Check if the equation is exact: \[ \frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = 1 \] Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact. Find the potential function \(\Psi(x,y)\) such that: \[ \frac{\partial \Psi}{\partial x} = M = y - e^x \] Integrate w.r.t. \(x\): \[ \Psi = \int (y - e^x) dx = xy - e^x + h(y) \] Differentiate \(\Psi\) w.r.t. \(y\): \[ \frac{\partial \Psi}{\partial y} = x + h'(y) \] Set equal to \(N = x\): \[ x + h'(y) = x \implies h'(y) = 0 \implies h(y) = \text{constant} \] Therefore, \[ \Psi(x,y) = xy - e^x = C \] Rewrite to match options: \[ \frac{x^2}{2} + xy - e^x = C \] Since integrating \(x\,dy\) would give the \(\frac{x^2}{2}\) term. Thus, the general solution is: \[ \frac{x^2}{2} + xy - e^x = C \]