# Video: Finding the Expression of a Function given Its Second Derivative Using Integration

Consider the initial value problem 𝑦′ = 4𝑥 − 3𝑦, 𝑦(0) = −1. Use Euler’s method with 𝑛 = 5 steps on the interval [0, 1] to find the value of 𝑦(1).

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### Video Transcript

Consider the initial value problem 𝑦 prime is four 𝑥 minus three 𝑦 where 𝑦 at 𝑥 equal to zero is negative one. Use Euler’s method with 𝑛 equal to five steps on the closed interval zero one to find the value of 𝑦 at 𝑥 is equal to one.

We’ve actually been given a differential equation 𝑦 prime is equal to four 𝑥 minus three 𝑦 and an initial value 𝑦 at 𝑥 is equal to zero is negative one. This means that our starting point for Euler’s method is the point 𝑥 zero, 𝑦 zero which is equal to zero, negative one. We also know that we have five steps and that our closed interval zero, one means that zero is less than or equal to 𝑥 is less than or equal to one. So we have 𝑛 equal to five and zero less than or equal to 𝑥 less than or equal to one.

We’re asked to use Euler’s method to find the value of 𝑦 at 𝑥 is equal to one. To do this, the first thing we need to do is to find the value of ℎ. That’s our step size. Now, ℎ is the interval width divided by the number of steps, in our case, the interval where there’s a maximum 𝑥 value one minus the minimum 𝑥 value zero which is one. The number of steps is five. So our step size is one over five. Let’s just sketch our Interval so that we can refer to it throughout our calculations. Our first 𝑥 value is zero, which is 𝑥 zero. Our step size is one over five, which is 0.2. So our next point 𝑥 one is 0.2. Our next 𝑥 value 𝑥 two is 0.4, 𝑥 three is 0.6, 𝑥 four is 0.8, and 𝑥 five is one.

And remember, we’re being asked to calculate the value of 𝑦 at 𝑥 is equal to one. So that’s 𝑦 of 𝑥 five, which is 𝑦 five. So we’re going to use Euler’s method to calculate 𝑦 five. We can’t go directly to 𝑦 five because the only value we have is 𝑥 zero, 𝑦 zero. So we start at this point and calculate 𝑦 one, 𝑦 two, then 𝑦 three, then 𝑦 four, and then 𝑦 five. So to use the formula, we need to know the value of 𝑥 zero, 𝑦 zero, which is zero, negative one. We need to know ℎ, which is 0.2, and we need to know what the function 𝑓 of 𝑥, 𝑦 is. In our differential equation, 𝑦 prime, which is d𝑦 by d𝑥, is equal to four 𝑥 minus three 𝑦. So 𝑓 of 𝑥 𝑦, in our case, is four 𝑥 minus three 𝑦.

We know that 𝑦 zero is equal to negative one. And we can use this to find 𝑦 one. And in the formula, if 𝑛 is equal to one, then 𝑛 minus one is equal to zero so that 𝑦 one is equal to 𝑦 zero plus ℎ times 𝑓 of 𝑥 zero, 𝑦 zero. We know that 𝑦 zero was negative one and that ℎ is 0.2. So we need to find 𝑓 of 𝑥 zero, 𝑦 zero. In our differential equation, 𝑓 of 𝑥 zero, 𝑦 zero is four 𝑥 zero minus three 𝑦 zero. And that’s equal to four times zero minus three times negative one, which is equal to three, so that 𝑦 one is equal to negative one plus 0.2 times three, which is negative 0.4. Now, we can use 𝑦 one to calculate 𝑦 two.

We know that 𝑦 one is equal to negative 0.4. We know that ℎ is 0.2. So we need to work out 𝑓 of 𝑥 one, 𝑦 one. And that’s equal to four 𝑥 one minus three 𝑦 one. And since 𝑥 one is 0.2, that’s four times 0.2 minus three times negative 0.4, which is 0.8 plus 1.2 which is two. So 𝑦 two is equal to negative 0.4 plus 0.2 times two which is actually zero. Now, if we put 𝑛 equal to three in our formula, 𝑦 three is equal to 𝑦 two plus ℎ times 𝑓 of 𝑥 two 𝑦 two. We know that 𝑦 two is equal to zero and ℎ is 0.2. So we need to work out 𝑓 of 𝑥 two 𝑦 two. And that’s equal to four 𝑥 two minus three 𝑦 two. And since 𝑥 two is 0.4, that’s four times 0.4 minus three times zero, which is 1.6. So we have 𝑦 three equal to zero plus 0.2 times 1.6, which is 0.32.

Now, with 𝑛 equal to four, we have 𝑦 four is equal to 𝑦 three plus ℎ times 𝑓 of 𝑥 three, 𝑦 three. 𝑓 of 𝑥 three 𝑦 three is four 𝑥 three minus three 𝑦 three. So we have four times 0.6 minus three times 0.32 which is 1.44, so that 𝑦 four is equal to 0.32 plus 0.2 times 1.44 and that’s 0.608. And finally with 𝑛 equal to five in our formula, we have 𝑦 five equal to 𝑦 four plus ℎ times 𝑓 of 𝑥 four 𝑦 four. 𝑓 of 𝑥 four 𝑦 four is equal to four 𝑥 four minus three 𝑦 four. And that’s equal to four times 0.8 minus three times 0.608 which is 1.376 so that we have 𝑦 five equal to 0.608 plus 0.2 times 1.376. And that’s equal to 0.8832. And remember that it’s exactly 𝑦 five that we’re looking for because this is the value of 𝑦 at 𝑥 is equal to one, so that 𝑦 of one is 0.8832.