Method of characterist PDE, transport equation with 3 variables

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Let $u=u(t,x,y)$. How can I use the method of the characteristis to solve the following transport equation:

$$\frac{\partial{u}}{\partial{t}}-\frac{\sinh(t)}{x}\frac{\partial{u}}{\partial{x}}-\frac{\cosh(t)}{y}\frac{\partial{u}}{\partial{y}}=\tan(t) $$

where $$ u(0,x,y)=x^2-y^2+1.$$

I don't know to solve this PDE because there is 3 variable. Anyone can shed light on how to solve this problem?

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$$\frac{\partial{u}}{\partial{t}}-\frac{\sinh(t)}{x}\frac{\partial{u}}{\partial{x}}-\frac{\cosh(t)}{y}\frac{\partial{u}}{\partial{y}}=\tan(t) $$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dt}{1}=\frac{-x}{\sinh(t)}dx=\frac{-y}{\cosh(t)}dy=\frac{du}{\tan(t)}$$ First characteristic equation from solving $\frac{dt}{1}=\frac{-x}{\sinh(t)}dx$ : $$x^2+2\cosh(t)=c_1$$ Second characteristic equation from solving $\frac{dt}{1}=\frac{-y}{\cosh(t)}dy$ : $$y^2+2\sinh(t)=c_2$$ Third characteristic equation from solving $\frac{dt}{1}=\frac{du}{\tan(t)}$ $$u+\ln|\cos(t)|=c_3$$ General solution of the PDE expressed on the form of implicit equation $\Phi(c_1\:,\:c_2\:,\:c_3)=0$ with arbitrary function $\Phi$ of three variables : $$\Phi\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\:,\:\left(u+\ln|\cos(t)|\right)\right)=0$$ Or equivalent $c_3=F(c_1\:,\:c_2)$ with arbitrary function $F$ of two variables : $$ \left(u+\ln|\cos(t)|\right)=F\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\right)$$ General solution : $$\boxed{u(t,x,y)=-\ln|\cos(t)|+F\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\right)}$$ INITIAL CONDITION : $u(0,x,y)=x^2-y^2+1$

$u(0,x,y)=-\ln|\cos(0)|+F\left(\left(x^2+2\cosh(0)\right)\:,\:\left(y^2+2\sinh(0)\right)\right)=F(x^2+2\:,\:y^2)$

$$u(0,x,y)=x^2-y^2+1=F(x^2+2\:,\:y^2)$$ Thus the function $F(X,Y)$ with $x^2=X-2$ and $y^2=Y$ must satisfy : $$F(X,Y)=(X-2)-Y+1=X-Y-1$$ Now the function $F(X,Y)$ is known. We put it into the above general solution where $X=x^2+2\cosh(t)$ and $Y=y^2+2\sinh(t)$.

$$u(t,x,y)=-\ln|\cos(t)|+x^2+2\cosh(t)-y^2-2\sinh(t)-1$$ This is the particular solution which satisfies both the PDE and the initial condition.