Question

Solve the following problems.

 

   \( t y \frac{d y}{d t}=\sqrt{t^{2} y^{2}+t^{2}+y^{2}+1} \)

Answer 1

Solution

Solve the following problems.

        \( t y \frac{d y}{d t}=\sqrt{t^{2} y^{2}+t^{2}+y^{2}+1} \)

\( \Longrightarrow t y \frac{d y}{d t}=\sqrt{t^{2}\left(y^{2}+1\right)+y^{2}+1} \)

\( \Longrightarrow t y \frac{d y}{d t}=\sqrt{\left(t^{2}+1\right)\left(y^{2}+1\right.}=\sqrt{t^{2}+1} \cdot \sqrt{y^{2}+1} \)

\( \Longrightarrow t y \frac{d y}{d t}=\sqrt{\left(t^{2}+1\right)\left(y^{2}+1\right.}=\sqrt{t^{2}+1} \cdot \sqrt{y^{2}+1} \)

\( \Longrightarrow \quad \frac{y}{\sqrt{y^{2}+1}} d y=\frac{\sqrt{t^{2}+1}}{t} d t \)

\( \Longrightarrow\int \frac{y}{\sqrt{y^2+1}}dy=\int \frac{\sqrt{t^{2}+1}}{t} d t \)

\( \Longrightarrow \sqrt{y^{2}+1}=\sqrt{t^{2}+1}-\sinh ^{-1}\left(\frac{1}{|t|}\right)+C \)

 Above integral show below:

Let \( u=y^{2}+1 \Longrightarrow d u=2 y d y \Longrightarrow d y=\frac{d u}{2} \)

\( \Longrightarrow \int \frac{y}{\sqrt{y^{2}+1}} d y=\int \frac{d u}{2 \sqrt{u}}=\sqrt{u}=\sqrt{y^{2}+1} \)

Let \( v=\sqrt{t^{2}+1} \Longrightarrow d v=\frac{t}{\sqrt{t^{2}+1}} d t \Longrightarrow d t=\frac{\sqrt{t^{2}+1}}{t} d v \)

\( \Longrightarrow \int \frac{\sqrt{t^{2}+1}}{t} d t=\int \frac{v^{2}}{v^{2}-1} d v=\int\left(\frac{u^{2}-1}{u^{2}-1}+\frac{1}{u^{2}-1}\right) d v \)

                          \( =\int\left(1+\frac{1}{u^{2}-1}\right) d v=v+\int\left(\frac{1}{(u-1)(u+1)}\right) d v \)

                          \( =v+\int\left(\frac{1}{2(u-1)}-\frac{1}{2(u+1)}\right) d v \)

                          \( =v+\frac{\ln (u-1)}{2}-\frac{\ln (u+1)}{2} \)

                          \( =\sqrt{t^{2}+1}+\frac{\ln \left(\sqrt{t^{2}+1}-1\right)}{2}-\frac{\ln \left(\sqrt{t^{2}+1}+1\right)}{2} \)

 

 

 

Therefore. \( \sqrt{y^{2}+1}=\sqrt{t^{2}+1}-\sinh ^{-1}\left(\frac{1}{|t|}\right)+C \)