Solving Separable Equations

Definition A first-order ODE is said to be separable if it can be put in the following form for some single variable (univariate) functions f (x) and g (y ), simply by factoring and simplifying. dy = f (x) · g (y ) dx If it does not have the above form then it is nonseparable. Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx = sin yx q = yx I dy dx = log I I = = y x x y
y
x Example Determine whether the following ODES are separable equations or nonseparable equations. I dy (nonseparable) dx = y + x I dy dx dy dx dy dx I dy dx = sin yx q = yx I dy dx = log I I = = y x x y
y
x Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx = sin yx q = yx I dy dx = log I I = = y x x y (separable)
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x Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx = sin x q = yx I dy dx = log I I = = y x x y (separable) y
y
x Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx = sin yx q = yx I dy dx = log I I = = y x x y
y
x (nonseparable) Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx I I I dy dx = = y x x y = sin yx q = yx = log
y
x (separable) Example Determine whether the following ODES are separable equations or nonseparable equations. I dy dx = y + x I dy dx dy dx dy dx I dy dx = sin yx q = yx I dy dx = log I I = = y x x y
y
x (nonseparable) A Note on the Substitution Method from Calculus 2 Why does the substitution method (i.e. u-substitution) work in Calculus 2? The idea is that if you choose u(x) carefully enough, then your integrand will just be f (u) · du dx . Effectively, what the substitution method tells you is that Z Z du f (u) · dx = f (u) du. dx Let’s see why that is... A Note on the Substitution Method from Calculus 2 Let’s define F (u) to be the antiderivative of f (u) (WITH R REPSECT TO u), i.e. F (u) = f (u) du. Then, this means dF = f (u) du by definition. Now, let’s compute dF dx by using the chain rule: dF du du dF = · = f (u) · . dx du dx dx R Thus, F (u) = f (u) · du dx dx as well. Essentially, what subsitution method says then is that Z Z du f (u) · dx = f (u) du. dx Method for Solving Separable Equations Consider the separable equation: dy dx 1 dy · g (y ) dx Z 1 dy · dx g (y ) dx Z 1 dy g (y ) G (y ) dy dx = f (x) · g (y ). = f (x) · g (y ) = f (x) Z = f (x) dx Z = f (x) dx = F (x) + C Provided you can integrate both the left hand side and the right hand side, you will have an equation relating just y and x. Definition An explicit solution to an ODE is a function y = f (x). This means that you can isolate y and solve for it completely in terms of x (i.e. y = x 2 ). An implicit solution to an ODE is an equation f (x, y ) = C . In this case it is impossible, or at least very challenging, to solve for y in terms of x, i.e. sin(y + x) = e xy + y . Example Find an explicit solution to the following IVP. dy = 2xy , y (0) = 3 dx Example Find an explicit solution to the following IVP. dy y −1 = , y (0) = 2 dx x +1 Example Find an explicit solution to the following IVP. dy = e 2x−3y , y (0) = 0 dx Example Find TWO explicit solutions to the following IVP. dy x(y 2 + 1) = , y (0) = 0 dx 2y Note: PLT does not apply here. Why? Example Find an implicit solution to the following IVP. dy = cos2 (y + 1), y (0) = π − 1 dx Note: PLT does apply here. What would the explicit solution be? Example Find an implicit solution to the following IVP. dy x +1 = y , y (0) = 0 dx e +y +3 Note: PLT does apply here. What would the explicit solution be? Hint: I DON’T KNOW THE EXPLICIT SOLUTION.