If the curve satisfying the differential equation dy/dx = 6-2e^(2x)y/1+e^(2x) passes through (0, 0) and (in 2, k), then k is

Question

If the curve satisfying the differential equation \(\frac{d y}{d x}=\frac{6-2 e^{2 x} y}{1+e^{2 x}}\) passes through (0, 0) and \((\ln 2, k),\) then k is

(1) \(\frac{3}{5} \ln 3\)

(2) \(\frac{6}{5} \ln 2\)

(3) \(\frac{8}{9} \ln 3\)

(4) \(\frac{7}{2} \ln 2\)

Answer 1

Correct option is (2) \(\frac{6}{5} \ln 2\)   

\(\frac{d y}{d x}=\frac{6-2 e^{2 x} y}{1+e^{2 x}}\)  

\(\frac{d y}{d x}=\left(\frac{2 e^{2 x}}{1+e^{2 x}}\right) y=\frac{6}{1+e^{2 x}}\)   

If \(=e^{\int \frac{2 e^{2 x}}{1+e^{2 x}} d x}\)   

\(=e^{\ln \left|1+e^{e x}\right|}=1+e^{2 x}\)