11–18. Solving initial value problems Use the method of your c...

Question

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(x) = x/y, y(2) = 4

Accepted Answer

Welcome back, everyone. Solve the initial value problem Y T equals Y divided by C and Y 1 equals 2 for T greater than 0. For this problem we're going to rewrite our left hand side as D Y divided by the T, that's our derivative in its differential form. So we get the Y divided by the T is equal toy divided by T. As we can notice, we can separate the variables and specifically what we're going to do is divide both sides by Y and multiply both sides by the T. So we get the Y divided by Y equals. DC divided by T. We have successfully separated the variables because we have Y terms on the left and C terms on the right. So let's go ahead and integrate both sides. On the left hand side we have LN of the absolute value of Y. On the right hand side we have LN of the absolute value of C. and we're going to add a constant of integration C for an indefinite integral. So now what we can do is solve for the absolute value of Y by taking E. To the power of the right hand side, so we get the absolute value of Y equals e to the power of Ln of the absolute value of T plus C. We know that T is greater than 0, so we can drop the absolute value for T, right? And now, because we have an exponential function on the right, specifically E raised an exponent, we know that this is always going to be greater than 0 as well, right? Because we have a positive leading coefficient in front of E, which is basically 1. So we can drop the absolute value for the absolute value of Y as well and simply rewrite it as Y equals. And now using the properties of exponents on the right hand side we get E to the power of LN of T multiplied by E raises to the power of C. Because we're going to have additional constants of integration, we can just label this constant as C1 for now. And now we can show that y equals e to the power of LN of T is simply t. And each the power of C1 can be written as some constant C2, right? So we get Y equals C2 multiplied by T. And now what we can do is simply solve the initial value problem, knowing that Y equals 2, if T is equal to 1. So we can now substitute Y equals 2, and we get 2 equals C2 multiplied by 1. And from here we can solve for C2, C2 is equal to 2 divided by 1, which is simply 2. And therefore our particular solution is Y equals C2 multiplied by T, which is 2T. That's our final answer and thank you for watching.

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.
y′(x) = x/y, y(2) = 4

Key Concepts

Separable Differential Equations

Initial Value Problems (IVP)

Integration Techniques

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