Question #4a5e2

3 Answers
Douglas K.
Apr 18, 2017

#y = -2/(x^2- 2)#

Explanation:

Given: #dy/dx = xy^2; y(0)=1#

Use the separation of variable method:

#dy/y^2= xdx#

Integrate:

#intdy/y^2= intxdx#

#-1/y = x^2/2+C#

#-2/y = x^2+C#

#y = -2/(x^2+ C)#

Evaluate at the boundary condition, #y(0) = 1#:

#1 = -2/(0^2+C)#

#C =-2#

The equation is:

#y = -2/(x^2- 2)#

salamat
Apr 18, 2017

#y = 2/(2 -x^2) #

Explanation:

#(dy)/(dx) = xy^2#

#1/y^2 dy= x dx#

#int 1/y^2 dy= int x dx#

#-1/y = x^2/2 +c#

plug in #y = 1 and x =0# in the above equation

#-1/1 = 0^2/2 +c# #-> c =-1#

therefore,
#-1/y = x^2/2 -1 = (x^2 - 2)/2#

#-2/(x^2 - 2) = y#

#2/(2 -x^2) = y#

Monzur R.
Apr 18, 2017

#y=2/(2-x^2)#

Explanation:

#dy/dx=xy^2#

#inty^-2# #dy=intx# #dx#

#-1/y=1/2x^2+"c"#

Given that #y=1, x=0#,

#-1/1=1/2(0)^2+"c" rArr"c"=-1#

Since they haven't asked to give it in the form #y=f(x)#, you can leave it like this. But I will solve it for #y# just in case you would like to see how to do that.

#-1/y=1/2x^2-1#

#1/y=1-1/2x^2=1/2(2-x^2)#

#y=(1/2(2-x^2))^-1#

#y=2/(2-x^2)#

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