Video: Solving a Differential Equation Involving Trigonometric Functions

Find the implicit solution to the following differential equation: sin (๐‘ฆ) (d๐‘ฆ/d๐‘ฅ) โˆ’ cos (๐‘ฅ) = 0.

03:02

Video Transcript

Find the implicit solution to the following differential equation. The sin of ๐‘ฆ multiplied by d๐‘ฆ by d๐‘ฅ minus the cos of ๐‘ฅ is equal to zero.

The question gives us a differential equation, and it wants us to find the implicit solution of this differential equation. Remember, when weโ€™re asked to find an implicit solution, we donโ€™t need to give our answer as a function of ๐‘ฅ. So letโ€™s try and find the solution to our differential equation. The first thing we notice is weโ€™re given a function of ๐‘ฆ and a function of ๐‘ฅ in our differential equation. Since thereโ€™s no way of manipulating our equation to remove our function in ๐‘ฆ, we should try writing this as a separable differential equation. In fact, in this case, we can just skip the step where we write this as a separable differential equation. Weโ€™ll just add the cos of ๐‘ฅ to both sides of the equation.

Doing this, we get the sin of ๐‘ฆ multiplied by d๐‘ฆ by d๐‘ฅ is equal to the cos of ๐‘ฅ. And now we want to separate our variables. We have our function of ๐‘ฆ on the left-hand side of our equation and our function of ๐‘ฅ on the right-hand side of our equation. And to solve this, we need to use a trick which we use when weโ€™re solving separable differential equations. We know d๐‘ฆ by d๐‘ฅ is not a fraction. However, we can treat it a little bit like a fraction when weโ€™re solving separable differential equations. This gives us the equivalent statement the sin of ๐‘ฆ d๐‘ฆ is equal to the cos of ๐‘ฅ d๐‘ฅ. And we can then solve this by integrating both sides of our equation. We get the integral of the sin of ๐‘ฆ with respect to ๐‘ฆ is equal to the integral of the cos of ๐‘ฅ with respect to ๐‘ฅ.

Weโ€™re now ready to evaluate both of these integrals. First, to integrate the sin of ๐‘ฆ with respect to ๐‘ฆ, we recall the integral of the sin of ๐œƒ with respect to ๐œƒ is equal to negative the cos of ๐œƒ plus a constant of integration. So by calling our variable ๐‘ฆ, we get that this integral was equal to negative the cos of ๐‘ฆ plus a constant of integration weโ€™ll call ๐ถ one. Now, to evaluate the integral of the cos of ๐‘ฅ with respect to ๐‘ฅ, we recall the integral of the cos of ๐œƒ with respect to ๐œƒ is equal to the sin of ๐œƒ plus a constant of integration. This time, since weโ€™re integrating the cos of ๐‘ฅ with respect to ๐‘ฅ, weโ€™ll label our variable ๐‘ฅ. This gives us the sin of ๐‘ฅ plus a constant of integration weโ€™ve called ๐ถ two.

At this point, we try and write this in the form ๐‘ฆ is equal to some function of ๐‘ฅ. But the question is telling us to find an implicit solution, so we donโ€™t need to do this step. Instead, we can rearrange our solution into any form which we want. Weโ€™ll start by combining our constants ๐ถ one and ๐ถ two into a new constant on the left-hand side of our equation weโ€™ll call ๐ถ. Next, weโ€™ll add the cos of ๐‘ฆ to both sides of this equation. This gives us ๐ถ is equal to the sin of ๐‘ฅ plus the cos of ๐‘ฆ. And weโ€™ll rearrange this equation one more time. Weโ€™ll rearrange this to get the cos of ๐‘ฆ plus the sin of ๐‘ฅ is equal to ๐ถ.

Therefore, given the differential equation the sin of ๐‘ฆ times d๐‘ฆ by d๐‘ฅ minus the cos of ๐‘ฅ is equal to zero, we were able to find the implicit solution the cos of ๐‘ฆ plus the sin of ๐‘ฅ is equal to ๐ถ.

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