Use the reduction of order method to find the general solutio of the following equations. One solution of the homogeneous is shown alongside each equation. $\newline$ $13$ . $2 x^{2} y^{\prime \prime}+3 x y^{\prime}-y=\frac{1}{x}, \quad y_{1}=x^{1 / 2}$ .

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Q. Use the reduction of order method to find the general solutio of the following equations. One solution of the homogeneous is shown alongside each equation.

\newline

13

2 x^{2} y^{\prime \prime}+3 x y^{\prime}-y=\frac{1}{x}, \quad y_{1}=x^{1 / 2}

Given Differential Equation: We are given the differential equation $2x^2y'' + 3xy' - y = \frac{1}{x}$ and one particular solution $y_1 = x^{1/2}$ . The reduction of order method involves finding a second solution of the form $y_2 = v(x)y_1$ , where $v(x)$ is a function to be determined.
Derivatives of y $2$ : First, we need to find the derivatives of $y_2$ . Since $y_2 = v(x)x^{1/2}$ , the first derivative is $y_2' = v'(x)x^{1/2} + \frac{1}{2}v(x)x^{-1/2}$ .
Substitute into Original Equation: Next, we find the second derivative of $y_2$ . This is $y_2'' = v''(x)x^{1/2} + v'(x)(1/2)x^{-1/2} - \frac{1}{4}v(x)x^{-3/2}$ .
Simplify and Combine Terms: Now we substitute $y_2$ , $y_2'$ , and $y_2''$ into the original differential equation in place of $y$ , $y'$ , and $y''$ respectively. This gives us: $\newline$ $2x^2(v''(x)x^{1/2} + v'(x)(1/2)x^{-1/2} - \frac{1}{4}v(x)x^{-3/2}) + 3x(v'(x)x^{1/2} + \frac{1}{2}v(x)x^{-1/2}) - v(x)x^{1/2} = \frac{1}{x}$ .
First-Order Linear Differential Equation: We simplify the equation by multiplying through and combining like terms. The terms involving $v(x)$ should cancel out because $y_1 = x^{1/2}$ is a solution to the homogeneous equation $2x^2y'' + 3xy' - y = 0$ . We are left with an equation involving only $v'(x)$ and $v''(x)$ .
Integrating Factor: After simplification, we should have an equation in the form $2x^2v''(x) + 2xv'(x) = \frac{1}{x}$ . This is a first-order linear differential equation in $v'(x)$ .
Multiplying by Integrating Factor: To solve for $v'(x)$ , we can use an integrating factor. The integrating factor is $e^{\int P(x) dx}$ , where $P(x) = \frac{1}{x}$ is the coefficient of $v'(x)$ in the simplified equation. Thus, the integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$ .
Integrate Both Sides: Multiplying the entire equation by the integrating factor $x$ , we get $2x^3v''(x) + 2x^2v'(x) = 1$ . This simplifies to $\frac{d}{dx}(x^2v'(x)) = 1$ .
Solve for v'(x): Integrating both sides with respect to $x$ , we get $x^2v'(x) = x + C_1$ , where $C_1$ is the constant of integration.
Integrate v'(x): Now we solve for $v'(x)$ by dividing both sides by $x^2$ , which gives us $v'(x) = \frac{1}{x} + \frac{C_1}{x^2}$ .
General Solution: We integrate $v'(x)$ to find $v(x)$ . This gives us $v(x) = \ln|x| - \frac{C_1}{x} + C_2$ , where $C_2$ is another constant of integration.
Substitute v(x) into y: The general solution to the differential equation is the sum of the particular solution and the multiple of the second solution. Therefore, the general solution is $y = C_1y_1 + y_2 = C_1x^{1/2} + v(x)x^{1/2}$ .
Simplify General Solution: Substituting $v(x)$ into the expression for $y$ , we get $y = C_1x^{1/2} + (\ln|x| - \frac{C_1}{x} + C_2)x^{1/2}$ .
Simplify General Solution: Substituting $v(x)$ into the expression for $y$ , we get $y = C_1x^{1/2} + (\ln|x| - \frac{C_1}{x} + C_2)x^{1/2}$ .Finally, we simplify the expression for $y$ to get the general solution in its simplest form: $y = C_1x^{1/2} + x^{1/2}\ln|x| - C_1x^{-1/2} + C_2x^{1/2}$ .