If #(1+x^2)dy/dx=x(1-y) and y(0)=4/3 #,then the value of #y(sqrt(8))+8/9# is ?

Question

If #(1+x^2)dy/dx=x(1-y) and y(0)=4/3 #,then the value of #y(sqrt(8))+8/9# is

After solving the differential equation I get
#ysqrt(1+x^2)=sqrt(1+x^2)+C#
what is to be done after this

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Answer 1 · 2017-12-17T12:17:54

#2#

You got

#y = 1+C_0/sqrt(x^2+1)#

Now, the condition #y(0) = 4/3 = 1+C_0/sqrt(0^2+1)# gives

#C_0 = 1/3# or #y = 1+1/3 1/sqrt(x^2+1)# and finally

#y(sqrt8)+8/9 = 1+1/3 1/sqrt(8+1) + 8/9=2#