In Exercises 1–22, solve the differential equation.

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Question

In Exercises 1–22, solve the differential equation.

xy' + 2y = 1 - x⁻¹

Accepted Answer

Welcome back everyone. Solve the differential equation xy minus Y equals 2x2 minus x to the power of -2, where x not equal to 0. For this problem, we're going to use the integrating factor and to use it, first of all, let's divide both sides of the equation by x to get y minus 1 divided by x multiplied by y. Equals 2 x minus x to the power of -3, right, so we have successfully divided both sides by x. Now we are going to identify the integrating factor mu which is e to the power of. Integral of -1 divided by XDX. And this is going to be equal to e to the power of negln of x, right? Using the properties of logarithms, we can show that this is e to the power of ln x to the power of -1. Which is X to the power of -1 or 1 divided by x. So what we want to do is now multiply both sides of our equation by the integrating factor which gives us 1 divided by x y minus 1 divided by x2. Y equals 2 x divided by x gives us 2 minus. X to the power of 3 divided by x gives us x to the power of -4. Now, let's notice that. On the left hand side we have the derivative of a product and specifically that's the derivative of y multiplied by 1 divided by x. This is equal to 2 minus x to the power of -4. And now from here. We are going to express the left hand side as d divided by dx. The derivative will be expressed in terms of the differential. Of y multiplied by 1 divided by x, which is equal to 2 minus x to the power of -4. If we multiply both sides by the x. We will be able to integrate. So now let's integrate the left-hand side and the right-hand side. On the left-hand side, we have the integral of the differential, which gives us why. Multiplied by 1 divided by X or basically Y divided by X. On the right hand side, the integral of 2DX is 2 X. And the integral of x, the power of -4, is going to be found using the power rule. So that's x to the power of -3. Divided by -3. Plus a constant of integration c. We can also simplify and show that this is 2 x plus. 1/3. X to the power of -3 + c. And now let's solve for y. We can multiply both sides by x, so that y is going to be. 2 X squared. Plus 1/3. X to the power of -3 + 1, that's x to the power of -2 + c x. That's our final answer, and thank you for watching.

In Exercises 1–22, solve the differential equation.xy' + 2y = 1 - x⁻¹

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In Exercises 1–22, solve the differential equation.

xy' + 2y = 1 - x⁻¹