Can someone solve this...xyy'=1-x^2?....thanks :)

2 Answers
James
May 6, 2018

answer
#y'=(1-x^2)/(x*y)#

Explanation:

i think that wanted
#xy*y'=1-x^2#

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

DanT
May 6, 2018

#y=sqrt(2lnx-x^2-c_1)#

Explanation:

First rewrite the differential equation. (Assume #y'# is just #dy/dx#):
#xydy/dx=1-x^2#

Next, separate the x's and y's- just divide both sides by #x# and multiply both sides by #dx# to get:
#ydy=(1-x^2)/xdx#

Now we can integrate both sides and solve for y:
#intydy=int(1-x^2)/xdx#
#intydy=int1/xdx-intx^2/xdx#
#y^2/2+c=lnx-intxdx#
(You only need to put the constant on one side because they cancel each other out into just one #c#.)

(Solving for y):
#y^2/2=lnx-x^2/2-c#
#y^2=2lnx-x^2-c_1# . (Can change to #c_1# after multiplying by 2)
#y=sqrt(2lnx-x^2-c_1)#

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