Question

Solve : xy' - y =e to the power y'. Also obtain its singular solution.

Answer 1

Answer on Question #40030 – Math – Differential Calculus

Solve: $xy' - y = e^{y'}$. Also obtain its singular solution.

Solution:

Solve the Clairaut equation $x \frac{dy(x)}{dx} - y(x) = e^{\frac{dy(x)}{dx}}$:

Subtract $x \frac{dy(x)}{dx}$ from both sides and divide by $-1$:

$$y(x) = -e^{\frac{dy(x)}{dx}} + x \frac{dy(x)}{dx}$$

Differentiate both sides with respect to $x$:

$$\frac{dy(x)}{dx} = \frac{dy(x)}{dx} - e^{\frac{dy(x)}{dx}} \frac{d^2 y(x)}{dx^2} + x \frac{d^2 y(x)}{dx^2}$$

Factor:

$$\frac{dy(x)}{dx} = \frac{dy(x)}{dx} + \frac{d^2 y(x)}{dx^2} \left(-e^{\frac{dy(x)}{dx}} + x\right)$$

Subtract $\frac{dy(x)}{dx}$ from both sides:

$$\frac{d^2 y(x)}{dx^2} \left(-e^{\frac{dy(x)}{dx}} + x\right) = 0$$

Solve $\frac{d^2 y(x)}{dx^2} = 0$ and $x - e^{\frac{dy(x)}{dx}} = 0$ separately:

For $\frac{d^2 y(x)}{dx^2} = 0$:

Integrate both sides with respect to $x$:

$$\frac{dy(x)}{dx} = \int 0 \, dx = c_1, \text{ where } c_1 \text{ is an arbitrary constant}$$

Substitute $\frac{dy(x)}{dx} = c_1$ into $y(x) = x\frac{dy(x)}{dx} - e^{\frac{dy(x)}{dx}}$ :

$$y (x) = - e ^ {c _ {1}} + c _ {1} x$$

For $x - e^{\frac{dy(x)}{dx}} = 0$ :

Solve for $\frac{dy(x)}{dx}$ :

$$\frac {d y (x)}{d x} = \log (x)$$

Substitute into $y(x) = x\frac{dy(x)}{dx} - e^{\frac{dy(x)}{dx}}$ :

$$y (x) = - x + x \log (x)$$

Collect solutions:

Answer:

$$y (x) = - e ^ {c _ {1}} + c _ {1} x \text{ or } y (x) = - x + x \log (x)$$