General solution to differential equation t^2 y' = 9 - y^2

[no1moparnut]

Find the general solution to t^2 y' = 9 - y^2
y = 2( 1 + Ce -4/t ) / ( 1 - Ce -4/t )
y = 2( 1 + Ce 4/t ) / ( 1 - Ce 4/t )
y = 3( 1 + Ce -6/t ) / ( 1 - Ce -6/t )
y = 3( 1 + Ce 6/t ) / ( 1 - Ce 6/t )
y = 4( 1 + Ce -8/t ) / ( 1 - Ce -8/t )
y = 4( 1 + Ce 8/t ) / ( 1 - Ce 8/t )
y = 5( 1 + Ce -10/t ) / ( 1 - Ce -10/t )
y = 5( 1 + Ce 10/t ) / ( 1 - Ce 10/t )
None of these.

 

[galactus]

To solve, separate variables.

\(\displaystyle t^{2}\cdot y'=9-y^{2}\)

\(\displaystyle \frac{y'}{9-y^{2}}=\frac{1}{t^{2}}\)

Integrate:

\(\displaystyle \frac{1}{6}ln\left(\frac{y+3}{y-3}\right)=\frac{-1}{t}+C\)

Continue?. Solve for y algebraically.