The differential equation given is:
y + 4y’ = 0
To find the general solution of this first-order linear ordinary differential equation, we can start by rearranging it:
4y’ = -y
Next, we can express this equation in a separable form, which involves isolating the variables. We divide both sides by 4:
y’ = -\frac{1}{4}y
Now, recall that y’ is the derivative of y with respect to x, which we can denote as
\frac{dy}{dx}:
\frac{dy}{dx} = -\frac{1}{4}y
To solve this separable equation, we can rearrange it to isolate the variables:
\frac{dy}{y} = -\frac{1}{4}dx
Next, we integrate both sides:
\int \frac{dy}{y} = -\frac{1}{4} \int dx
Performing the integration gives us:
\ln |y| = -\frac{1}{4}x + C
Where C is the constant of integration. We can exponentiate both sides to eliminate the natural logarithm:
|y| = e^{(-\frac{1}{4}x + C)} = e^C e^{-\frac{1}{4}x}
Letting K = e^C, where K is a positive constant, we get:
|y| = K e^{-\frac{1}{4}x}
Since y can take both positive and negative values, we can drop the absolute value to write the general solution:
y = Ke^{-\frac{1}{4}x}
Here, K represents any constant, which accounts for the generality of the solution. Thus, the final answer is:
y = Ke^{-\frac{1}{4}x}
In summary, the general solution of the differential equation y + 4y’ = 0 is:
y(x) = Ke^{-\frac{1}{4}x}