Solve the initial-value problem. $x y ^ { t } - y = x \ln x , y ( 1 ) = 3$
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

This is a first-order linear differential equation of the form $xy' - y = x\ln(x)$. To solve it, we use an integrating factor, which in this case is $e^{\int -1/x \,dx} = e^{-\ln(x)} = 1/x$. Multiplying both sides of the equation by this integrating factor transforms it into $\frac{d}{dx}(y/x) = \ln(x)$. Integrating both sides with respect to $x$ gives $y/x = \frac{1}{2}(\ln(x))^2 + C$, where $C$ is the constant of integration. Multiplying both sides by $x$ gives $y = \frac{1}{2}x(\ln(x))^2 + Cx$. Using the initial condition $y(1) = 3$, we find that $C = 3$, so the solution is $y(x) = \frac{1}{2}x(\ln(x))^2 + 3x$.

Solve the Initial-Value Problem xyt−y=xln⁡x,y(1)=3x y ^ { t } - y = x \ln x , y ( 1 ) = 3xyt−y=xlnx,y(1)=3

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