Initial Value Problems - At A Glance

Answer

Let's check the initial condition first:

y(0) = 4e⁰ = 4,

so the initial condition is satisfied. Now for the differential equation. The right-hand side of the differential equation is 0. We have

y' = -4e^-x

and

y" = 4e^-x,

so the left-hand side of the d.e. is

y" + y' = 4e^-x + (-4e^-x) = 0.

Since the left- and right-hand sides of the d.e. come out the same, the function y = 4e^-x satisfies the differential equation. Since the function y = 4e^-x satisfies both the d.e. and the initial condition, this function is a solution to the IVP.

Answer

Let's check the initial condition first:

f (0) = (0)³ ≠ 2.

Since the initial condition fails, f (x) = x³ is not a solution to the IVP.

Answer

The initial condition holds:

y(0) = (0) + 2 = 2.

The left-hand side of the d.e. is

2y = 2(x + 2) = 2x + 4.

Since the derivative of y is 1, the right-hand side of the d.e. is

Since the two sides of the d.e. are not equal, the function y = x + 2 is not a solution to the d.e. and therefore can't be a solution to the IVP either.

Answer

The initial condition is satisfied, since sin (0) = 0. We need some derivatives to check the differential equation:

f '(x) = cos x

f ⁽²⁾(x) = -sin x

f ⁽³⁾(x) = -cos x.

The left-hand side of the d.e. is

f ⁽³⁾(x) = -cos x.

The right-hand side of the d.e. is

-f '(x) = -cos x.

Since the two sides of the d.e. agree, the function f (x) = sin x is a solution to the differential equation. Since the function f (x) = sin x satisfies both the d.e. and the initial condition, this function is a solution to the IVP.

Answer

The initial condition is satisfied:

y(1) = (1)² + 4(1) = 5.

We have

and

The left-hand side of the d.e. is

The right-hand side of the d.e. is

Since the two sides of the d.e. are the same, this function is a solution to the differential equation. Since this function also satisfies the initial condition, the function y = x² + 4x is a solution to the IVP.