The function y = 4x + 1 satisfies the differential equation, since

However,

y(0) = 4(0) + 1 = 1

so the function y = 4x + 1 doesn't satisfy the initial condition. This means the function y = 4x + 1 doesn't satisfy the IVP.

Example 2

Find a solution to the IVP

We know that the equation

y = 4x + 1

doesn't work because when x = 0 we get y = 1 instead of y = 2. We still want a line with slope 4, but we want it to be 2 when x = 0. This sounds like the line

y = 4x + 2

is a good bet. Let's check and see if this works. The derivative is

so this function satisfies the d.e. part. When x = 0 we get

y = 4(0) + 2 = 2,

so this function also satisfies the initial condition. That means

y = 4x + 2

is a solution to this initial value problem.

It's usually easier to check if the function satisfies the initial condition(s) than it is to check if the function satisfies the d.e., so we recommend checking the initial condition(s) first.

Example 3

Determine if the function y = 5x^{2} + 3x satisfies the IVP

y" = 10 and y(0) = 3.

Since

y(0) = 5(0)^{2} + 3(0) = 0 ≠ 3,

this function doesn't satisfy the IVP. Since the function didn't satisfy the initial condition, we don't need to bother checking if the function satisfied the differential equation.