Exact solution at t = 2 and x = 1 (PDE) 
I need help in this question. I did the following steps:   
(a) I considered the characteristic $dx/dt = \frac{3t}{3x^2 +5}$ and solve it using the separation method and   obtained $x^3 + 5x = 3/2t^2 + c$ . Then I considered (xo, to)  to obtain that $x^3 + 5x = 3/2t^2 + xo^3 + 5xo - 3/2to^2$  
(b) Then along the characteristic $du/dt = \frac{-tU}{x^3 +5x +1}$. Plugging in $x^3 + 5x = 3/2t^2 + xo^3 + 5xo - 3/2to^2$ and using the separation method I end up with the following   
$U(x,t) = K(3/2t^2 + x0 + 5x0 - 3/2to^2 +1)^-1/3$   
(c) When I plug x = 1 and t = 2, I have that the final solution for $U(x,T)$ at that value is $U(1,2) = \frac{1}{5^1/3} = .5848$
 
But I have the feeling it is not correct. 
 A: Begin by noting that we have PDE:
$$ u_t +\frac{3t}{3x^2+5}  u_x  = - \frac{t}{x^3 + 5x + 1}u$$
Divide through by $u$ to yield
$$ \frac{u_t}{u} +\frac{3t}{3x^2+5}  \frac{u_x}{u} = - \frac{t}{x^3 + 5x + 1}$$
Observe that $\frac{u_t}{u} = \frac{\partial}{\partial t} \ln (u)$ and $\frac{u_x}{u} = \frac{\partial}{\partial x} \ln (u)$. So let $g = \ln (u)$  then the problem becomes:
$$ g_t +\frac{3t}{3x^2+5} g_x = - \frac{t}{x^3 + 5x + 1}$$
And this reduces to
$$ (3x^2 + 5)g_t +3t g_x = - \frac{3tx^2 + 5t}{x^3 + 5x + 1}$$
This is in standard form for attack! since $A(x) g_t + B(t) g_x=0$ has solution $$ F\left( \int B(t) dt - \int A(x) dx \right)$$ and the general solution to $A(x) g_t + B(t) g_x = E(x,y)$ is a single solution to the latter diff EQ PLUS ANY solution to the homogenous case (ask in comments if this isn't clear). 
So we wish to find a single solution $\mu$ to:
$$ (3x^2 + 5)g_t +3t g_x = - \frac{3tx^2 + 5t}{x^3 + 5x + 1}$$
And the know that $\mu + F \left( \frac{3}{2}t^2 - x^3 - 5x \right)$ is the general solution for any $F$. We now go backwards (you'll see why in a minute!) to
$$ \frac{1}{3t} g_t + \frac{1}{3x^2 + 5}g_x = - \frac{1}{x^3 + 5x+1}$$
Let $g_t = 0$ (since the right side is entirely dependent on x we can do this)  then we have that $g_x = - \frac{3x^2 + 5}{x^3 + 5x + 1}$ which can be integrated with our old friend partial fractions (or more slickly notice that $3x^2 + 5 = \frac{d}{dx}(x^3 + 5x + 1)$) giving us a solution:
$$ \mu = \ln(x^3 + 5x + 1)$$
So the general solution then will be:
$$ g = \mu + F\left(\frac{3}{2}t^2 - x^3 - 5x \right) \rightarrow $$
$$ g = \ln(x^3 + 5x + 1) + F\left(\frac{3}{2}t^2 - x^3 - 5x \right)$$
But $g = \ln u$ so we have that $u = e^g$, thus
$$ u = e^{\ln(x^3 + 5x + 1) + F\left(\frac{3}{2}t^2 - x^3 - 5x \right)}= (x^3 + 5x + 1)F\left(\frac{3}{2}t^2 - x^3 - 5x \right)$$
(the $F$ absorbs the exponent, and the natural logarithm disappears). 
Now things get ugly... It's easy to satisfy $u(0,t) = cos(4t)$ as
$$ u = (x^3 + 5x + 1) \cos \left(\sqrt{16t^2 - \frac{32}{3}x^3 - \frac{160}{3}x} \right) $$
But the other condition doesn't have a straightforward fix.
