Video: Finding the Equation of a Curve given Its Tangent in Terms of π‘₯ and 𝑦 by Solving a Separable Differential Equation

Find the equation of the curve that passes through the point (0, βˆ’1) given d𝑦/dπ‘₯ = (βˆ’6π‘₯ βˆ’ 4)/(4𝑦 + 13).

03:24

Video Transcript

Find the equation of the curve that passes through the point zero, negative one given d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four divided by four 𝑦 plus 13.

For this question, we first noticed that we have been given a separable differential equation. This is an equation that can be written in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. Solving this separable differential equation will give us a general solution, which will include a constant 𝑐. Now, the question has also given us an initial value, which is the fact that the curve should pass through the point zero, negative one. We’ll be able to use this information to narrow down the general solution that we find into a specific solution. For now, let’s just work on the general solution.

For now, we’ll put aside our initial value and just work towards the general solution. To find this, we’ll first express our differential equation in the following form, where we have some function of 𝑦 d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯. For our equation, we can do this by multiplying both sides by four 𝑦 plus 13. This gives us that four 𝑦 plus 13 d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four. The next thing we can do is to treat d𝑦 by dπ‘₯ somewhat like a fraction. If we pretend we’re multiplying both sides by dπ‘₯, we obtain an equivalent statement four 𝑦 plus 13 d𝑦 is equal to negative six π‘₯ minus four dπ‘₯.

The next step we take is to then integrate both sides of our equation, giving us an integral with respect to 𝑦 on the left-hand side and an integral with respect to π‘₯ on the right-hand side. Now, you should always remember that what we have done here is essentially a trick. d𝑦 by dπ‘₯ is not actually a fraction and cannot always be treated like one. However, we’ve done so here to save us from working since it gives us an equivalent answer. Going back to our method, we now perform our integrations. Doing so gives us the following.

Now, it’s worth noting that both of our integrals would actually have given us a constant. However, we can simply combine these two constants into one term. Instead of 𝑐 one and 𝑐 two, we simply have a 𝑐, which is equal to 𝑐 one minus 𝑐 two. Again, this leads to less unnecessary work for us. After some small simplifications, we have arrived at the general solution to our differential equation. This means that we have found an equation which describes all the lines for which d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four divided by four 𝑦 plus 13.

However, we need to find which one of these lines passes through the point that we’ve been given: zero, negative one. This initial value that we have been given will allow us to find that particular solution. We know that if the line passes through the point zero, negative one that when π‘₯ is equal to zero, 𝑦 must be equal to negative one. We can, therefore, take our general solution, substitute in 𝑦 equals negative one and π‘₯ equals zero, and solve this to find the value of our constant 𝑐. With a little bit of working, we find that the value of 𝑐 here is negative 11. We can now substitute this value of 𝑐 back into our general solution to find the specific solution. This is the specific solution to the differential equation which passes through the point zero, negative one, which is our initial value.

Now, you might notice that the solution that we have been given is implicit, which means it’s not in the form 𝑦 equals some function of only π‘₯. Usually, it’s a good idea to try and express your solution in explicit form. However, in this case, you should not worry. It is not actually possible to give the answer to this problem in the form of a single explicit equation since we have a multiple powers of 𝑦 on the left-hand side. Again, this is no cause for concern, since the implicit form is a perfectly valid solution. And we have indeed answered our question.

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