### Video Transcript

Suppose that dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Find π¦ in terms of π₯.

The question gives us a first-order differential equation. It wants us to find π¦ in terms of π₯ given that when π¦ is equal to seven π₯ is equal to π by six. We recall if weβre told the derivative of π¦ with respect to π₯ is equal to some function of π₯, we can solve for π¦ by doing the opposite of differentiating. We have that π¦ is equal to the integral of dπ¦ by dπ₯ with respect to π₯, which we know is equal to the integral of negative nine times the sin of two π₯ minus three times the cos of five π₯ with respect to π₯.

To evaluate this integral, we can integrate each term separately. And we recall for constants π and π where π is not equal to zero, the integral of π cos of ππ₯ with respect to π₯ is equal to π times the sin of ππ₯ divided by π plus a constant of integration πΆ. And the integral of π sin of ππ₯ with respect to π₯ is equal to negative π the cos of ππ₯ divided by π plus the constant of integration πΆ.

Using this, we get the integral of negative nine times the sin of two π₯ is negative one multiplied by negative nine cos of two π₯ divided by two. And negative one multiplied by negative one is just equal to one. So, integrating the first term, we get nine cos of two π₯ divided by two. We can then do the same to evaluate the integral of negative three times the cos of five π₯. We get negative three times the sin of five π₯ divided by five. And then, we add our constant of integration πΆ. And itβs worth noting we only need to add one constant of integration.

So, weβve now shown that π¦ is equal to this function of π₯, where πΆ can be any constant. However, weβre told that when π¦ is equal to seven, π₯ is equal to π by six. We can use this information to find the specific value of πΆ in this case. We substitute π¦ is equal to seven and π₯ is equal to π by six into this equation. This gives us that seven is equal to nine times the cos of two times π by six over two minus three times the sin of five times π by six over five plus πΆ.

We can simplify this. We have two times π by six is equal to π by three. Weβre now in a position to evaluate this expression. The cos of π by three is a standard trigonometric result which we should know. Itβs equal to one-half. And the sin of five π by six is also a standard trigonometric result we should know. Itβs also equal to one-half. This gives us that seven is equal to nine times one-half over two minus three times one-half over five plus πΆ. We can simplify this further. Nine times one-half over two is the same as saying nine over four. And three times one-half over five is the same as saying three over 10.

So, we have that seven is equal to nine over four minus three over 10 plus πΆ. We can rearrange this expression and then evaluate to get that πΆ is equal to 101 divided by 20. We then substitute this value of πΆ into our general solution. And this gives us that π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20. Therefore, weβve shown if dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Then π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20.