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Solve the differential equation \frac{dy}{dx} + 2xy = 0
Answer: The differential equation given is:
\frac{dy}{dx} + 2xy = 0
This is a first-order linear differential equation. To solve this, we can use the method of integrating factors.
Step 1: Identify the Integrating Factor
The general form of a first-order linear differential equation is:
\frac{dy}{dx} + P(x)y = Q(x)
Here, P(x) = 2x and Q(x) = 0 .
The integrating factor \mu(x) is given by:
\mu(x) = e^{\int P(x) \, dx} = e^{\int 2x \, dx} = e^{x^2}
Step 2: Multiply the Differential Equation by the Integrating Factor
Multiply both sides of the differential equation by e^{x^2} :
e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y = 0
Step 3: Simplify and Integrate
Notice that the left side of the equation is the result of the product rule:
\frac{d}{dx} \left( y e^{x^2} \right) = 0
Integrate both sides with respect to x :
\int \frac{d}{dx} \left( y e^{x^2} \right) \, dx = \int 0 \, dx
This simplifies to:
y e^{x^2} = C
where C is the constant of integration.
Step 4: Solve for y
Finally, solve for y :
y = C e^{-x^2}
So, the general solution to the differential equation \frac{dy}{dx} + 2xy = 0 is:
y = C e^{-x^2}