Differential Equation Exercise 4
We’re given the following first-order differential equation:
$$ xy' + y = 0 $$
This is a first-order differential equation because the highest derivative that appears is the first derivative, y'.
Let’s isolate y' in terms of x and y:
$$ y' = - \frac{y}{x} $$
This is a separable differential equation of the form y' = f(x)g(y), with f(x) = - 1/x and g(y) = y.
We now rewrite the equation using the notation dy/dx:
$$ \frac{dy}{dx} = - \frac{y}{x} $$
Next, we separate the variables:
$$ \frac{dy}{y} = - \frac{dx}{x} $$
We integrate both sides with respect to their respective variables:
$$ \int \frac{dy}{y} = \int - \frac{dx}{x} $$
$$ \int \frac{1}{y} \, dy = - \int \frac{1}{x} \, dx $$
The integral on the left yields the antiderivative F(y) = log(y) + c1:
$$ \log(y) + c_1 = - \int \frac{1}{x} \, dx $$
Evaluating the right-hand side gives F(x) = log(x) + c2:
$$ \log(y) + c_1 = -\log(x) + c_2 $$
We can combine the constants c1 and c2 into a single constant c3 = c2 + c1.
The sign of the constant is irrelevant, as arbitrary constants can take any real value.
$$ \log(y) = -\log(x) + c_3 $$
Bringing the logarithmic terms to the same side:
$$ \log(y) + \log(x) = c_3 $$
Using the logarithmic identity log(y) + log(x) = log(xy):
$$ \log(xy) = c_3 $$
Exponentiating both sides to eliminate the logarithm:
$$ e^{\log(xy)} = e^{c_3} $$
$$ xy = e^{c_3} $$
Since \( e^{c_3} \) is just a positive constant, we can denote it as c:
$$ xy = c $$
Solving for y, we find the general solution of the differential equation:
$$ y = \frac{c}{x} $$
Alternative Solution Method
We can also solve the original differential equation using a different approach:
$$ xy' + y = 0 $$
Divide both sides of the equation by x:
$$ \frac{xy' + y}{x} = \frac{0}{x} $$
$$ y' + \frac{y}{x} = 0 $$
This is a linear first-order differential equation of the form y' + a(x)y = b(x), where a(x) = 1/x and b(x) = 0.
We can solve it using the integrating factor method:
$$ y = e^{-\int a(x) \, dx} \cdot \left[ \int b(x) \cdot e^{\int a(x) \, dx} \, dx + c \right] $$
Substituting a(x) = 1/x and b(x) = 0 gives:
$$ y = e^{-\int \frac{1}{x} \, dx} \cdot \left[ \int 0 \cdot e^{\int \frac{1}{x} \, dx} \, dx + c \right] $$
$$ y = e^{-\int \frac{1}{x} \, dx} \cdot c $$
The integral ∫1/x dx = log(x) + constant, but since we already have a constant c outside, we write:
$$ y = e^{-\log(x)} \cdot c $$
$$ y = \frac{1}{e^{\log(x)}} \cdot c $$
Because the exponential and the logarithm are inverse functions, e{log(x)} = x:
$$ y = \frac{1}{x} \cdot c $$
Once again, we find the general solution of the differential equation:
$$ y = \frac{c}{x} $$
As expected, both methods lead to the same result.
And so on.
