Question

Solve the following problems.

\( (2 t-y-1) d y=(3 t+y-4) d t \)

Answer 1

Solution

Solve the following problems.

\( (2 t-y-1) d y=(3 t+y-4) d t \)

\( \Longrightarrow \frac{d y}{d t}=\frac{3 t+y-4}{2 t-y-1} \quad\text {where, }\left|\begin{array}{cc} 3 & 1 \\ 2 & -1 \end{array}\right|=-5 \neq 0 \)

Choose a, b s.t.\( \left\{\begin{array}{l}3 a+b-4=0 \\ 2 a-b-1=0\end{array} \Longrightarrow a=1, b=1\right. \)

Let's \( t=u+a=u+1, y=v+b=v+1 \)

\( \Longrightarrow d t=d u, d y=d v \Longrightarrow \frac{d y}{d t}=\frac{d v}{d u} \)

\( \Longrightarrow \frac{d v}{d u}=\frac{3(u+1)+(v+1)-4}{2(u+1)-(v+1)-1}=\frac{3 u+v}{2 u-v}=\frac{3+v / u}{2-v / u} \)

Then, let's \( k=\frac{v}{u} \Longrightarrow v=k u \Longrightarrow \frac{d v}{d u}=k+u \frac{d k}{d u} \)

\( \Longrightarrow k+u \frac{d k}{d u}=\frac{3+k}{2-k} \Longrightarrow u \frac{d k}{d u}=\frac{k^{2}-k+3}{2-k} \)

\( \Longrightarrow \frac{1}{u} d u=\frac{2-k}{k^{2}-k+3} d k \)

\( \Longrightarrow \ln (u)=\frac{3 \sqrt{11}}{11} \arctan \left(\frac{2 k-1}{11}\right)-\frac{1}{2} \ln \left(k^{2}-k+3\right)+C \)

where \( k=\frac{v}{u}, u=t-1, v=y-1 \Longrightarrow k=\frac{y-1}{t-1} \)

 

Therefore.

\( \ln (u)=\frac{3 \sqrt{11}}{11} \arctan \left(\frac{2 k-1}{11}\right)-\frac{1}{2} \ln \left(k^{2}-k+3\right)+C \)