Solve the differential equation in Exercises 9–22.
15. √...

Question

Solve the differential equation in Exercises 9–22.

15. √x (dy/dx) = e^(y+√x), x > 0

Accepted Answer

Welcome back everyone. Solve the differential equation 2 square root of x. Dy divided by dx equals e to the power of y plus 2 square root of x for x greater than 0 and express the solution explicitly for y. For this problem we are going to use separation of variables. What we can do is rewrite the equation as follows. So first of all, we have two square root of x multiplied by the derivative of y with respect to x on the left hand side. On the right hand side we will use the properties of exponents, so we can separate the exponents e to the power of y multiplied by e to the power of 2 square root of x. Now from here. Let's multiply both sides by dx so that 2 square root of x, Dy is equal to e to the power of y, e to the power of 2 square root of x dx. And now what we can do is essentially divide both sides by the square root of x, so let's go ahead and do that. Which essentially eliminates square root of x on the left hand side. And we can also divide both sides by e to the power of y, right? So that'd be 2 divided by e to the power of y Dy equals e to the power of 2 square root of x d x divided by square root of x. Now, from here. What we can do is essentially. Rewrite the left hand side as 2e to the power of negative YDY. And on the right hand side we have e to the power of 2 square root of x, the x divided by the square root of x. Since we have separated the variables successfully. We can now Integrate both sides, so we're going to show that we're integrating the left hand side and the right hand side. For the left hand side, the integral of 2 e to the power of negydy can be calculated as 2 integral of e to the power of negative ydy, which is 2 multiplied by e to the power of negative Y divided by -1. So we basically get the negative sign in front, right? Plus a constant of integration, let's call it C1. Now for the other integral, we are going to use U substitution. Let's suppose that U is equal to the exponent 2 square root of x. The derivative of u with respect to x is going to be 2 multiplied by the derivative of the square root of x, which is 1 divided by 2 square root of x. And this is equal to 1 divided by a square root of x. If we multiply both sides by dx, we can show that du is equal to dx divided by the square root of x. So considering the integral. E e to the power of 2 square root of x, d x divided by the square root of x can now be rewritten as. The integral of e to the power of u multiplied by the u. And this integral is equal to e to the power of u plus a constant of integration c2, which is e to the power of 2 square root of x plus c2. Well then, so now we have our results for the integrals. Let's rewrite them. We basically get -2 e to the power of y plus C1 equals. E to the power of 2 square root of x. Plus C2. What we can do from here is combine the constants of integration into C on the right-hand side. And from here, we can divide both sides by 2. So for now, let's just keep C in terms of a new constant C3, because we are still dividing by another constant. And dividing both sides by -2. We get E raise the power of negative Y. Equals 1/22. E So the power of 2 square root of x plus. C3 divided by -2, right? This is how we can express it for now. And essentially. To solve for Y, we're going to take the natural log of both sides. We can reassign our constant C3 divided by -2 to some constant C. And let's take Ellen off both sides. Which will give us negative y, right? y equals ln of. We can now rewrite our terms, starting with the positive term C minus 1/2. E to the power of 2 square root of x. And finally, we can multiply both sides by -1 to get y. Y equals ln of C minus 1/2 e to the power of 2 square root of x. That's our final answer for this problem. Thank you for watching.

Solve the differential equation in Exercises 9–22.15. √x (dy/dx) = e^(y+√x), x > 0

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Solve the differential equation in Exercises 9–22.
15. √x (dy/dx) = e^(y+√x), x > 0