Questions
a
t=gxx+c
b
t=gxx2+c
c
t=gxx+c
d
t=gx+x+c
detailed solution
Correct option is C
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=tTalk to our academic expert!
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