Consider the differential equation ( \frac{dy}{dx}=2xy ). This first‑order ordinary differential equation appears frequently in mathematics, physics, and engineering because it models processes where the rate of change of a quantity is proportional to both the quantity itself and the independent variable. And understanding how to solve it, interpret its solutions, and apply the results builds a solid foundation for tackling more complex differential equations. In this article we will walk through the theory behind the equation, derive its general and particular solutions, examine the qualitative behavior of those solutions, and illustrate practical situations where the equation arises Simple as that..
Understanding the Equation
The notation ( \frac{dy}{dx}=2xy ) states that the derivative of (y) with respect to (x) equals the product of (2), the independent variable (x), and the dependent variable (y). On the flip side, because the right‑hand side can be factored into a function of (x) times a function of (y) (namely (2x) and (y)), the equation is separable. Separability is a key property that allows us to isolate all (y)-terms on one side of the equation and all (x)-terms on the other, making integration straightforward Still holds up..
A few immediate observations help set the stage:
- If (y=0) for any (x), then (\frac{dy}{dx}=0); thus the constant function (y(x)=0) is a solution (the trivial solution).
- For any non‑zero initial value, the solution will never cross the (x)-axis because the differential equation preserves the sign of (y).
- The factor (2x) changes sign at (x=0); consequently, the direction of the slope field reverses when moving from negative to positive (x).
Solving the Differential Equation
Separation of Variables
Starting from [ \frac{dy}{dx}=2xy, ]
we divide both sides by (y) (assuming (y\neq0)) and multiply by (dx):
[ \frac{1}{y},dy = 2x,dx. ]
Now each side depends on only one variable, so we can integrate:
[ \int \frac{1}{y},dy = \int 2x,dx. ]
The left integral yields (\ln|y|) (the natural logarithm of the absolute value of (y)), while the right integral gives (x^{2}+C), where (C) is an arbitrary constant of integration. Thus,
[ \ln|y| = x^{2}+C. ]
Exponentiating both sides eliminates the logarithm:
[ |y| = e^{x^{2}+C}=e^{C}e^{x^{2}}. ]
Since (e^{C}) is always positive, we can replace it with a new constant (K>0) and drop the absolute value by allowing (K) to take any real value (positive, negative, or zero). This yields the general solution
[ \boxed{y(x)=Ke^{x^{2}}}, ]
where (K) is an arbitrary real constant. Notice that the trivial solution (y=0) corresponds to (K=0) And that's really what it comes down to. Turns out it matters..
Verification
Differentiating (y=Ke^{x^{2}}) gives
[ \frac{dy}{dx}=K\cdot e^{x^{2}}\cdot 2x = 2x\bigl(Ke^{x^{2}}\bigr)=2xy, ]
confirming that the expression satisfies the original differential equation for any constant (K).
Particular Solutions and Initial Conditions
To obtain a specific solution we need an initial condition, typically given as (y(x_{0})=y_{0}). Substituting into the general solution:
[ y_{0}=Ke^{x_{0}^{2}} \quad\Longrightarrow\quad K = y_{0}e^{-x_{0}^{2}}. ]
Hence the particular solution satisfying the condition is
[y(x)=y_{0}e^{-x_{0}^{2}}e^{x^{2}} = y_{0}e^{,x^{2}-x_{0}^{2}}. ]
Examples
| Initial condition | Constant (K) | Particular solution |
|---|---|---|
| (y(0)=1) | (K=1) | (y=e^{x^{2}}) |
| (y(1)=2) | (K=2e^{-1}) | (y=2e^{x^{2}-1}) |
| (y(-2)=0) | (K=0) | (y=0) (trivial) |
These examples illustrate how the sign and magnitude of (K) dictate whether the solution grows, decays, or remains identically zero The details matter here..
Qualitative Behavior of Solutions
Growth Rate
Because the exponential term (e^{x^{2}}) increases faster than any ordinary exponential (e^{ax}) (for fixed (a)) as (|x|\to\infty), solutions with non‑zero (K) exhibit super‑exponential growth in both the positive and negative (x) directions. The factor (x^{2}) ensures symmetry: (y(x)) depends only on (x^{2}), so the solution is an even function when (K\neq0).
- For (x>0), as (x) increases, (x^{2}) grows, making (y) increase dramatically.
- For (x<0), the same increase occurs because (x^{2}) is unchanged; thus the solution mirrors itself across the (y)-axis.
Sign Preservation
If (K>0) (equivalently (y_{0}>0)), then (y(x)>0) for all (x). If (K<0) ((y_{0}<0)), then (y(x)<0) everywhere. The solution never crosses zero unless it is identically zero, a direct consequence of the uniqueness theorem for first‑order ODEs when the function (f(x,y)=2xy) and its partial derivative (\partial f/\partial y = 2x) are continuous Less friction, more output..
Slope Field Insight
A quick sketch of the slope field shows arrows pointing away from the (x)-axis as (|x|) grows, with the slope magnitude increasing with both (|x|) and (|y|). Near the origin, the slope is small because the factor (2x) is near zero; this explains why solutions appear relatively flat close to (x=0) before accelerating outward.
Applications Although the equation (dy/dx=2xy) is simple, it appears in several modeling contexts:
-
Population Dynamics with Quadratic Time Dependence
In certain biological models, the per‑capita growth rate may increase linearly with time (e.g., due to improving resources). Setting the growth rate proportional to both the current population and time leads exactly to this differential equation. -
Physics: Charged Particle in a Time‑Varying Magnetic Field The radial component of the velocity of a charged particle moving in a magnetic
