Question

In Exercises 23–28, solve the initial value problem.x dy/dx + 2y = x² + 1, x \u003e 0, y(1) = 1

Accepted Answer

Rewrite the given differential equation in the standard linear form. Start by dividing the entire equation by $$x ($$since $$x \u003e 0) to $$isolate $$\frac{dy}{dx}$$:

$$\frac{dy}{dx} + \frac{2}{x} y = x + \frac{1}{x}$$ Identify the integrating factor $$\mu(x)$$, which is given by

$$\mu(x) = e$$^{\int P(x) \, dx}$$, where P(x$$) = \frac{2}{x}$$.  
Calculate the integral:

$$\int \frac{2}{x} \, dx = 2 \ln|x| = \ln|x|^2$$

So,

$$\mu(x) = $$e^{\ln|x|^2}$$ = $$x^2 ($$since $$x \u003e 0). $$Multiply both sides of the differential equation by the integrating factor $$x^2$$:

$$x^2$$ \frac{dy}{dx} + 2x y = $$x^3 + x$

Notice that the left side is the derivative of the product x^2$ y$$, so rewrite it as:

$$\frac{d}{dx} (x^2$ y) = x^3$ + x$$ Integrate both sides with respect to $$x$$:

$$\int \frac{d}{dx} (x^2$ y) \, dx = \int (x^3$ + x) \, dx$$

This simplifies to:

$$x^2 y$$ = \int $$x^3$$ \, dx + \int x \, dx + C$$

Calculate the integrals on the right side. Solve for y$ by dividing both sides by x^2$:

y$$ = \frac{1}{$$x^2$$} \left( 	ext{integrated expression} + C ight)$$

Finally, use the initial condition y(1) = 1$ to find the constant C$ by substituting x=1$ and y=1$ into the equation.

In Exercises 23–28, solve the initial value problem.
x dy/dx + 2y = x² + 1, x > 0, y(1) = 1

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Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Initial Value Problems (IVP)

In Exercises 23–28, solve the initial value problem.x dy/dx + 2y = x² + 1, x > 0, y(1) = 1