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# Question Video: Solving a Separable First-Order Differential Equation Mathematics • Higher Education

Solve the differential equation π¦β² + π₯π^(π¦) = 0.

04:38

### Video Transcript

Solve the differential equation π¦ prime plus π₯ times π to the power of π¦ is equal zero.

Weβre given a differential equation, and weβre asked to find the solution to this differential equation. The first thing we should do is try and write our differential equation in a form which we know how to solve. So to start, weβll write π¦ prime as dπ¦ by dπ₯. So weβve rewritten our differential equation as dπ¦ by dπ₯ plus π₯ times π to the power of π¦ is equal to zero. The next thing we could try and do is write our differential equation in the form dπ¦ by dπ₯ is equal to some function of π₯. Then we could just integrate both sides of the equation with respect to π₯.

However, we have a problem. We have π to the power of π¦ in our differential equation. And we canβt remove this. So weβll always going to have this function of π¦ in our differential equation. We can see that itβs multiplied by π₯. This should remind us of separable differential equations. Remember, we call a differential equation separable if it can be written in the form dπ¦ by dπ₯ is equal to some function π of π¦ times some function π of π₯. In our case, we can do this by subtracting π₯ times π to the power of π¦ from both sides of our differential equation. This gives us our function π of π¦ is π to the power of π¦ and our function π of π₯ is negative π₯. And the reason we call these separable differential equations is we can try and solve these by separating our variables.

To separate our variables, weβll start by dividing both sides of the equation through by π to the power of π¦. Using our laws of exponents, weβll write this as π to the power of negative π¦ times dπ¦ by dπ₯ is equal to negative π₯. Now, remember, dπ¦ by dπ₯ is not a fraction. However, when weβre solving separable differential equations, we can treat it a little bit like a fraction. This gives us the equivalent statement π to the power of negative π¦ dπ¦ is equal to negative π₯ dπ₯. Then all we need to do is integrate both sides of our equation. We get the integral of π to the power of negative π¦ with respect to π¦ is equal to the integral of negative π₯ with respect to π₯. We can now evaluate both of these integrals.

Letβs start with the integral of π to the power of negative π¦ with respect to π¦. We recall, for any constant π not equal to zero, the integral of π to the power of ππ¦ with respect to π¦ is equal to one over π π to the power of ππ¦ plus the constant of integration πΆ. We just divide by the coefficient of π¦ in our exponent. In our case, the coefficient of π¦ is negative one. So we want to divide our integrand by negative one. This gives us negative π to the power of negative π¦. And we could add a constant of integration here. However, remember, weβre also going to integrate negative π₯ with respect to π₯. So weβll just combine these two constants of integration together.

Now, we want to evaluate the integral of negative π₯ with respect to π₯. Remember, negative π₯ is actually negative π₯ to the first power. So we can integrate this by using the power rule for integration, which tells us, for any constant π not equal to negative one, the integral of π₯ to the πth power with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus the constant of integration πΆ.

We want to add one to our exponent of π₯ and then divide by this new exponent. In our case, the exponent of π₯ is one. So we add one to this to get two and then divide by two. This gives us negative π₯ squared divided by two plus the constant of integration weβll call πΆ one.

And this is a general solution to our differential equation. However, this is an implicit solution because itβs not of the form π¦ is equal to some function of π₯. So if possible, we should try to write this in the form π¦ is equal to some function of π₯. So letβs try and do this. Weβll start by multiplying both sides of our equation through by negative one. And remember, πΆ one is a constant. So we can write negative πΆ one as a new constant weβll call πΆ. So we now have π to the power of negative π¦ is equal to π₯ squared over two plus πΆ.

Remember, we want to rewrite this equation so π¦ is the subject. So weβll need to take the natural logarithm of both sides of our equation. Doing this, we get the natural logarithm of π to the power of negative π¦ is equal to the natural logarithm of π₯ squared over two plus πΆ. Now, remember, the natural logarithm function and the exponential function are inverses. So the natural logarithm of π to the power of negative π¦ would just be equal to negative π¦. So we now have negative π¦ is equal to the natural logarithm of π₯ squared over two plus πΆ. And the last thing we need to do is multiply both sides of this equation through by negative one.

And this means weβve written our general solution in the form π¦ is equal to some function of π₯. Therefore, given the differential equation, π¦ prime plus π₯ times π to the power of π¦ is equal to zero, we were able to find the solution π¦ is equal to negative one times the natural logarithm of π₯ squared over two plus πΆ.

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