Question

solve the differential equation:(1+y²) (1+ logx)dx + xdy=0, given that y=1​

Answer 1

Answer:

Step-by-step explanation:

We have

(1+y²)(1+logx)dx+xdy=0

⇒(1+logx)dx/x+1dy/(1+y²)=0 [on separating the variables]

⇒∫(1+logx)dx/x+∫1dy(1+y²)=C [integrating both sides]

⇒∫tdt+tan⁻¹y=C, where (1+logx)=t

⇒1/2t²+tan⁻¹y=C, where C is an arbitrary constant

⇒1(1+logx)²/2+tan⁻¹y=C.  ...(i)

Putting x=1andy=1 in (i), we get

C=1/2+tan⁻¹1⇒C=(1/2+π/4). ...(ii)

∴1/2(1+logx)²+tan⁻¹y=(1/2+π/4) [using (ii) in (i)]

⇒1/2(logx)²=logx+tan⁻¹y=π/4, which is the required solution