1. Find the solution of the initial-value problem: x^ay'+2x^a-1y=x^b,y(1)=1 with xgt 0 where a,b are real numbers.

## Step 1: Rewrite the equation<br />### Divide the equation by $x^a$ to get it in standard form: $y' + \frac{2}{x}y = x^{b-a}$.<br /><br />## Step 2: Find the integrating factor<br />### The integrating factor is $e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = e^{\ln(x^2)} = x^2$.<br /><br />## Step 3: Multiply by the integrating factor<br />### Multiply the equation by $x^2$: $x^2y' + 2xy = x^{b-a+2}$.<br /><br />## Step 4: Integrate both sides<br />### Notice the left side is the derivative of $x^2y$. Integrating both sides with respect to $x$ gives $\int (x^2y' + 2xy) dx = \int x^{b-a+2} dx$, which simplifies to $x^2y = \frac{x^{b-a+3}}{b-a+3} + C$ assuming $b-a+3 \neq 0$.<br /><br />## Step 5: Solve for y<br />### Divide by $x^2$ to solve for $y$: $y = \frac{x^{b-a+1}}{b-a+3} + \frac{C}{x^2}$.<br /><br />## Step 6: Apply the initial condition<br />### Substitute $y(1) = 1$: $1 = \frac{1^{b-a+1}}{b-a+3} + \frac{C}{1^2}$, so $1 = \frac{1}{b-a+3} + C$. Thus, $C = 1 - \frac{1}{b-a+3} = \frac{b-a+2}{b-a+3}$.<br /><br />## Step 7: Final Solution<br />### Substitute the value of C back into the equation for y: $y = \frac{x^{b-a+1}}{b-a+3} + \frac{b-a+2}{b-a+3}\frac{1}{x^2}$ for $b-a+3 \neq 0$.<br />### If $b-a+3=0$, then $\int x^{b-a+2}dx = \int x^{-1}dx = \ln|x|+C$. So $x^2y = \ln|x|+C$. Since $x>0$, $x^2y = \ln x + C$. Using $y(1)=1$, $1 = \ln 1 + C$, so $C=1$. Thus $y = \frac{\ln x + 1}{x^2}$.