Way back in algebra we learned that a solution to an equation is a value of the variable that makes the equation true. This is backwards kind of thinking we need for differential equations.
To check if a number is a solution to an equation, we evaluate the left-hand side of the equation at that number, then evaluate the right-hand side of the equation at that number. If we get the same value for both sides of the equation, the number is a solution to the equation.
Example 1
Determine if x = 5 is a solution to the equation x2 + 6 = 4x + 11. |
Example 2
Determine if x = 4 is a solution to the equation x2 + 6 = 4x + 11. |
Example 3
Determine whether the function y = x2 is a solution to the d.e. xy' = 2y. |
Example 4
Determine whether the function y = x2 + 4 satisfies the d.e. xy' = 2y. |
Example 5
Show that the function y = xex is a solution to the d.e. |
Example 6
Show that y = x ln x is not a solution to the d.e. |
Exercise 1
Determine whether y = ex is a solution to the d.e.
y' + y" = 2y.
Exercise 2
Determine whether y = xex is a solution to the d.e.
y' = xy.
Exercise 3
Determine whether P = e-t is a solution to the d.e.
Exercise 4
Determine whether y = x2 is a solution to the d.e.
Exercise 5
Determine whether y = xn is a solution to the d.e.
x2y" + ny = n2y.
Exercise 6
Show that y = 5x2 + 2 is a solution to the d.e.
Exercise 7
Show that y = 5x2 is not a solution to the d.e.
Exercise 8
Show that f (x) = sin x is a solution to the d.e.
f (2)(x) + f (x) = 0.
Exercise 9
Show that f (x) = exsin x is a solution to the d.e.
2f '(x) – 2f (x) = f ''(x).
Exercise 10
Show that y = ln x is a solution to the d.e.
ey = x2y'.