Home » Nonlinear Differential Equations

# Find solutions of the Riccati equation y′ + y + y2 = 2

Consider the Riccati equation

This equation has two constant solutions. Starting with these and the previous exercise (linked above) find further solutions in the cases:

1. If , find a solution on the interval with when .
2. If or , find a solution on the interval with when .

First, we find the two constant solutions. If is constant then so

From Exercise 19 (linked above) we know we can obtain additional solutions to the Riccati equation by

where is a solution of

In the present case we have , so is the solution of either

for the cases and , respectively. Each of these is an first-order linear differential equation which we can solve using Theorem 8.3 (page 310 of Apostol). For the first one we have , and let , and . Then we have

This gives us the first solution

Evaluating the second differential equation, this time with , , and we have,

Therefore, the second solution is

Finally, for the specific cases in (a) and (b).

1. We want , so we choose . Then . (This follows since .) Therefore, , so,

where .

2. In this case we want or , so we choose . Since we have . Therefore,

where .

# Prove a formula for additional solutions of a Riccati equation given one solution

Consider a differential equation

called a Riccati equation. Prove that if is a solution of this equation then there are additional solutions

where satisfies a first-order linear differential equation.

Proof. Let be a function satisfying

Further, let where satisfies a first-order linear differential equation. Then,

Therefore,

Thus, if

then is a solution of our equation. But,

This is a first-order linear differential equation (since is a function of ). Hence, is a solution of

where is the solution of the first-order linear differential equation

# Find solutions of 2xyy′ + (1+x)y2 = ex for given initial values

Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

1. when ;
2. when ;
3. a finite limit as .

From a previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

1. In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

2. From part (a) we have which implies . Since we then have

3. Let where is some finite real number. Then, (using the calculations we already completed in part (a) and just changing the initial-values and )

Then,

Therefore,

# Find all solutions of xy′ + y = y2x2 log x for given initial values

Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

From a previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

# Find solutions of xy′ – 2y = 4x3y1/2 for given initial values

Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

From a previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

# Find solutions of y′ -y = -y2(x2 + x + 1) for given initial values

Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

From a previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

# Find solutions of y′ – 4y = 2exy1/2 for given initial values

Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

From the previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

# Find all solutions of a given initial value problem

Let the function be the unique solution of the initial-value problem

with where is a constant, and are continuous functions on an interval , and with any real number. If and prove that the function for which for all is a solution of

if and only if for all .

Proof. Assume is a solution of . Let . Then,

(This division is allowed since on implies on .) Therefore,

Thus, if is a solution of then is the unique solution of

Conversely, if on then

Therefore,

Then, since we know by hypothesis that is the unique solution of we have

Substituting this into the above equation (and noting that our assumption that implies ) we have

Therefore is a solution of with