Question 2.5.4: Solve the differential equation t/y y+ln(ty)+1 = 0

We begin by observing that this equation is neither linear nor separable. Thus, writing the derivative in differential notation, we have

\frac {t}{y} \frac {dy}{dt}+ln(ty)+1 = 0

and then rearranging algebraically,

(ln(ty)+1)dt + \frac {t}{y} dy = 0        (2.5.10)

Letting M(t , y) = ln(ty)+1 and N(t , y) = t /y, we observe that

M_{y} =\frac {1}{ty} t=\frac {1}{y}   and   N_{t} =\frac {1}{y}

and therefore, M_{y} = N_{t} . Hence the differential equation is exact and we can assume that a function φ exists such that φ_{t} =M(t , y)   and   φ_{y} = N(t , y).

the latter equation is more elementary, we consider φ_{y} = t /y, and integrate both sides with respect to y. Doing so, we find that

φ(t , y) = t ln y +h(t)       (2.5.11)

From (2.5.10), φ must also satisfy φ_{t} =ln(ty)+1, so we take the partial derivative of both sides of (2.5.11) with respect to t to find that

φ_{t} = ln y +h´(t ) = ln(ty)+1

From this and properties of the logarithm, we observe that

ln y +h´(t ) = ln t +ln y +1

and thus h´(t ) = ln t +1. It follows (integrating by parts and simplifying) that h(t ) = t ln t . Thus, we have demonstrated that the original equation is indeed exact by finding φ(t , y)=t ln y +t ln t =t ln(ty). From here, we now know that φ(t , y) = K, and so

t ln(ty) = K

Solving for y, we have that

y =\frac {1}{t} e^{K/t}