Find a particular solution to the IVP

y' = 4 and y(1) = 9.

We know any solution to the d.e.

y' = 4

has the form

y = 4x + C.

If we're to have y(1) = 9, then

9 = 4(1) + C

and so C has to equal 5. There's no choice. This means the specific solution to this IVP is

y = 4x + 5.

The usual process for solving IVPs is to

• get a general solution by solving the d.e., then

• get a particular solution by using the initial condition(s) to find the constants.

We'll be able to solve more kinds of differential equations later, but we can do some practice ones now by thinking backwards.

Find the particular solution to the IVP

y' = 6x and y(0) = 3.

The general solution to the d.e. y' = 6x is

y = 3x² + C

Convince yourself: take the derivative of y = 3x² + C. For the particular solution, we want to have y = 3 when x = 0, so we can plug these numbers into the general solution and solve for C:

3 = 3(0)² + C

so C = 3. The particular solution to this IVP is

y = 3x² + 3.

Solve the IVP

y" = 12x and y'(0) = 5 and y(0) = 4

We need to think backwards twice to find the general solution. Thinking backwards from y" to y', we get

y' = 6x² + B.

Thinking backwards again, we see that y must look like this:

y = 2x³ + Bx + C.

To find the particular solution, let's first use the condition y(0) = 4. Plugging x = 0 and y = 4 into our equation for y gives us

4 = 2(0)³ + B(0) + C

so C = 4. To find B we take the condition y'(0) = 5 and plug x = 0 and y' = 5 into the equation for y':

5 = 6(0)² + B

so B = 5. The particular solution to this IVP is

y = 2x³ + 5x + 4.

In the previous example we had a second-order differential equation (link to defn of second-order) and two initial conditions. For a first-order differential equation we need only one initial condition. A differential equation of order n will have a general solution with n undetermined constants, and you need n initial conditions to figure out a particular solution.

Solve the IVP

f ⁽³⁾(x) = 0, f (0) = 0, f '(0) = 1, f "(0) = 4.

First we find the general solution. If the third derivative of f is 0, the second derivative of f must be a constant:

f ⁽²⁾(x) = A.

The first derivative must be a linear function with slope A:

f '(x) = Ax + B.

The original function must look like this:

Now we use the initial conditions to fill in the constants. When x = 0 we want f(0) = 0, so

which means C = 0. When x = 0 we want f ' (0) = 1, so

1 = A(0) + B

which means B = 1.

When x = 0 we want f "(0) = 4, so

4 = A.

Putting everything we know about the constants together, we get

Of course, we can check our answer by taking 3 derivatives of f to make sure that f ⁽³⁾(x) = 0 (it does) and making sure that f, f ', and f " have the correct values when x = 0.