Solve the initial-value problem.
$x y ^ { \prime } - y = x \ln x , \quad y ( 1 ) = 3$
A) $y ( x ) = 3 x ( \ln x ) ^ { 2 }$
B)
$y ( x ) = \frac { 1 } { 2 } x ( \ln x ) ^ { 2 } + 3 x$
C)
$y ( x ) = 3 x \ln x + \frac { 1 } { 2 } x$
D)
$y ( x ) = 3 \ln x + \frac { 1 } { 2 } x$
E)
$y ( x ) = \frac { 1 } { 2 } ( \ln x ) ^ { 3 } + x$

Question

Quizplus · Accepted Answer

The integrating factor is $x$. Multiplying both sides by $x$, we get$$x^2y' - xy = x^2\ln x$$$$(x^2y)' = x^2\ln x$$Integrating both sides, we get$$x^2y = \frac{1}{2}x^2(\ln x)^2 + C$$$$y = \frac{1}{2}x(\ln x)^2 + Cx^{-1}$$Using the initial condition $y(1) = 3$, we get$$3 = \frac{1}{2}(1)(\ln 1)^2 + C(1)^{-1}$$$$3 = \frac{1}{2} + C$$$$C = \frac{5}{2}$$Therefore, the solution to the initial-value problem is$$y = \frac{1}{2}x(\ln x)^2 + \frac{5}{2x}$$

Solve the Initial-Value Problem xy′−y=xln⁡x,y(1)=3x y ^ { \prime } - y = x \ln x , \quad y ( 1 ) = 3xy′−y=xlnx,y(1)=3

Solve the Initial-Value Problem $x y ^ { \prime } - y = x \ln x , \quad y ( 1 ) = 3$