Question

The substitution \[ \frac{dy}{dx}=z, \] reduces the differential equation \[ \frac{d^2y}{dx^2}-\frac{dy}{dx}=0 \] to a differential equation whose solution is \(z=\)

• A. \(\log x+C\)
• B. \(x+C\)
• C. \(Ae^x\)
• D. \(x^2+C\)

Hint:

Whenever \[ \frac{dy}{dx}=z \] is substituted, remember: \[ \frac{d^2y}{dx^2}=\frac{dz}{dx}. \] This converts a second-order differential equation into a first-order equation.

Answer 1

The Correct Option is C

Step 1: Use the given substitution.
Given: \[ \frac{dy}{dx}=z \] Therefore, \[ \frac{d^2y}{dx^2}=\frac{dz}{dx} \] Substitute into the differential equation: \[ \frac{d^2y}{dx^2}-\frac{dy}{dx}=0 \] We get \[ \frac{dz}{dx}-z=0 \]

Step 2: Solve the first-order differential equation.
Rearranging, \[ \frac{dz}{dx}=z \] So, \[ \frac{dz}{z}=dx \] Integrating both sides: \[ \int \frac{1}{z}\,dz=\int dx \] \[ \log z=x+C \] Exponentiating, \[ z=e^{x+C} \] \[ z=Ae^x \] where \[ A=e^C \]

Step 3: Final conclusion.
Hence, \[ \boxed{z=Ae^x} \]