Solution of differential equation dtdx=tddx(g(x))−t2g(x) is

Rearranging the terms of equation,  We get dtdx−tg′(x)g(x)=t2g(x)⇒1t2dtdx+1tg′(x)g(x)=1g(x)…………..(1) Let z=1t⇒−1t2dtdx=dzdx  Thus, from (i) we obtain dzdx+g′(x)g(x)z=1g(x) Which if clearly linear in 2 and dzdx with I.F=e∫g′(x)g(x)dx=elog⁡[g(x)]=g(x)⇒ Thus complete solution is z. g(x)=∫g(x)⋅1g(x)dx+c⇒1tgx=x+c  ⇒  gxx+c=t