Differential Equations Exercises

Answer

First we find the general solution. If y' = 4x³ then

y = x⁴ + C.

Since we're given the initial condition that y = 20 when x = 2, we have

20 = (2)⁴ + C = 16 + C.

This means C = 4 and so the particular solution is

y = x⁴ + 4.

Answer

The general solution is

y = cos x + C.

Using the initial condition y(0) = 2, we have

2 = cos 0 + C = 1 + C

so C = 1. The particular solution is

y = cos x + 1.

Answer

The general solution is

Using the initial condition that says y = 3 when x = 1, we have

Solving, we get and so the particular solution is

Answer

We need to think backwards twice. The first time, we get

y' = 2x² + B.

The second time, we get

When x = 0 we want y to be 7, which means

so C = 7. When x = 0 we want y' to be 5, so we get

5 = 2(0)² + B

and so B = 5. The particular solution is

Answer

The general solution is

y = -e^x + C.

Using the initial condition, we have

-3 = -e⁰ + C = -1 + C.

We conclude that C = -2 and so the particular solution is

y = -e^x - 2.