Question

Verify that y = log x, is a solution of the differential
dy
equation
dx
2
+dy by dx
= 0​

Answer 1

Answer:

Given, (1−y  

2

)(1+logx)dx+2xydy=0

⇒  

x

(1+logx)dx

​  

=  

(1−y  

2

)

2ydy

​  

 

Integrating, we get ∫  

x

(1+logx)dx

​  

=∫  

(1−y  

2

)

2ydy

​  

 

Let logx=t, so  

x

dx

​  

=dt

Also, let y  

2

=s, so 2ydy=ds

⇒∫(1+t)dt=∫  

1−s

ds

​  

 

⇒t+  

2

t  

2

 

​  

=ln(1−s)+c

⇒logx+  

2

(logx)  

2

 

​  

=ln(1−y  

2

)+c

When y=0,x=1; we get 0+0=0+c

⇒c=0

⇒logx+  

2

(logx)  

2

 

​  

=ln(1−y  

2

)

Step-by-step explanation:

Answer 2

y = log x + c

Differentiating both sides w.r.t.x, we get

dydx=1x+0=1xdydx=1x+0=1x

xdydx=1xdydx=1

Differentiating againi w.r.t x, we get

xd2ydx2+dydx×1=0∴xd2ydx2+dydx=0xd2ydx2+dydx×1=0∴xd2ydx2+dydx=0

The shows that y= log x +c is a solution of the D.E . xd2ydx2+dydx=0