How to Solve a Differential Equation with Separation of Variables
3 Examples Solving a Separable Differential Equation
A differential equation of the form
y' = f(x)g(y)
is called a separable differential equation because it is possible to separate the variables and transform the equation into the form A(y) = B(x). If the function A(y) is invertible and of a suitable form, one can express y as a function of x. Here are several examples to show how the method works.
Example 1Solve the differential equation y' = x2ey. First write y' as dy/dx and treat the derivative as a fraction. By cross multiplying dy/dx = x2ey, we arrive at
e-y dy = x2 dx.
Now integrate both sides:
-e-y = (1/3)x3 + c.
It is necessary to append the "+ c" to the right hand side because different initial conditions yield different final solutions. Since the left hand side is invertible, we can solve for y:
y = -ln[-(1/3)x3 - c].
If we are supplied with the initial condition y(0) = 0.3, then we find that c = -0.7408. So the final solution is
y = -ln[0.7408 - (1/3)x3].
Example 2Solve dy/dx = x2ey/y. Following the method above gives us
ye-y dy = x2 dx,
-ye-y - e-y = (1/3)x3 + c.
Unfortunately, there is no way to solve the left-hand side for y, so this is the most simplified form of the solution.
Example 3Solve y' = tan(y/x) + y/x. On the surface, this does not look separable because we can't untangle the tangent function. However, we can use change of variables with the substitution y = zx and y' = z'x + z. Now the problem becomes
z'x + z = tan(z) + z
z'x = tan(z)
x dz/dx = tan(z)
cot(z) = 1/x dx
ln(sin(z)) = ln(x) + c
sin(z) = ecx
z = arcsin[ecx]
Using the relation y = zx, the solution is y = (x)arcsin[ecx]. With the initial condition y(1) = 1, we find that c = -0.1726, so the final solution is y = (x)arcsin[0.8415x].
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