Question

In exercises 1–4, verify that the given formula is a solution to the initial value problem.

2. Powers of t.

b) y ′ = t^3 , y(0) = 5: y(t) = (1/5)t^(4) + 5

3. Sines and cosines.

a) x′ = −y, y′ = x, x(0) = 1, y(0) = 0: x(t) = cost, y(t) = sint

Answer 1

Step 1)

we know that if function is a solution to the initial value problem than it must satisfies the differential equation and initial condition

we have y' = t³ and y(0) = 5

we have,

we can say that y(t) is a solution of the differential equation of it satisfies the differential equation y' = t³

we can write,

--------------------------------------------------------1)

Let's assume that y = (1/5)t⁴ + 5 hence we can say that,

Put it in equation 1) we can say that,

Step 2)

There might be a typo error while posting the question and the formula for y(t) should be,

we can write,

--------------------------------------------------------2)

Let's assume that y = (1/4)t⁴ + 5 hence we can say that,

Put it in equation 2) we can say that,

we can see that y(t) = (1/4)t⁴ + 5 satisfies the differential equation hence it is a solution to given differential equation

now we will check whether y(t) satisfies the initial condition or not

we have y(0) = 5

we have y(t) = (1/4)t⁴ + 5 hence put t = 0 we can write,

Hence we can say that y(t) = (1/4)t⁴ + 5 also satisfies the initial condition

As y(t) = (1/4)t⁴ + 5 satisfies both differential equation and initial condition we can say that y(t) = (1/4)t⁴ + 5 is a solution to given initial value problem

Note :

As y(t) = (1/5)t⁴ + 5 does not satisfy differential equation there is no need to check for initial condition and directly we can say that y(t) = (1/5)t⁴ + 5 is not a solution to initial value problem

Step 3)

we have,

x(t) = cost and y(t) = sin(t)

Hence let's assume that x = cost and y = sint

Also differential equation is given by x' = -y and y' = x hence we can write,

-----------------------------------------a)

------------------------------------------b)

we have x = cost and y = sint hence we can write,

Put x' = -sint and y = sint in equation a) we can write,

Hence we can say that x = cost and y = sint satisfies the differential equation x' = -y

we have x = cost and y = sint hence we can write,

Put y' = cost and x = cost in equation b) we can write,

Hence we can say that x = cost and y = sint satisfies the differential equation y' = x

Now we will check for the initial condition

we have x(t) = cos(t) and y(t) = sin(t) also we have x(0) = 1 and y(0) = 0

Put x = 0 and y = 0 we can write,

Hence we can say that x(t) = cost and y(t) = sint satisfies the initial conditon

As x(t) = cost and y(t) = sint satisfies both differential equations and initial conditions we can say that x(t) = cost and y(t) = sint is a solution of the initial value problem