The general solution of the differential equation d y d x = ^2(3 x+y) is ⁻¹ ( 3 2 (3 x+y) )=f(x) . Then, f(x)=

Step-by-step solution

Here, d y d x = ^2(3 x+y) On putting 3 x+y=t3+ d y d x = d t d x array ll & d y d x = d t d x -3 d t d x -3= ^2 t & d t d x = ^2 t+3 d t ^2 t+3 =d x array Integrate both side, d t ^2 t+3 = d x aligned & ^2 t d t 1+3 ^2 t = d x ^2 t 1+3+3 ^2 t = d x & ^2 t d t 4+3 ^2 t = d x aligned On putting t=m , ^2 t d x=d m= 1 4 d m 1+ ( 3 2 m )^2 x+C= 1 4 2 3 ⁻¹ ( 3 2 m )=x+C= 1 2 3 ⁻¹ [ 3 2 t ]=x+C[ m= t]= ⁻¹ [ 3 2 (3 x+y)=2 3 (x+C) ][ t=3 x+y] So, f(x)=2 3 (x+C)