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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 1
Solve the differential equation y' = x^{2}e^{y}. First write y' as dy/dx and treat the derivative as a fraction. By cross multiplying dy/dx = x^{2}e^{y}, we arrive ate^{-y} dy = x^{2} dx.
Now integrate both sides:
-e^{-y} = (1/3)x^{3} + 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)x^{3} - 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)x^{3}].
Example 2
Solve dy/dx = x^{2}e^{y}/y. Following the method above gives usye^{-y} dy = x^{2} dx,
-ye^{-y} - e^{-y} = (1/3)x^{3} + 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 3
Solve 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 becomesz'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) = e^{c}x
z = arcsin[e^{c}x]
Using the relation y = zx, the solution is y = (x)arcsin[e^{c}x]. 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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