subject to the limits: y = 1 x = 0
the answer is: y = 1 + 2x
however i cannot get this answer, here is my attempt
separating the variables:
x + 1 dy = y + 1 dx
integrating both sides:
x^2/2 +x = y^2/2 + x
inputting values and solving for c:
c = -1/2
after putting the value of c back in and rearranging i get:
y = √x^2 + x + 1/2
clearly i am doing something wrong. please help show how to get the correct answer!!!
the answer is: y = 1 + 2x
however i cannot get this answer, here is my attempt
separating the variables:
x + 1 dy = y + 1 dx
integrating both sides:
x^2/2 +x = y^2/2 + x
inputting values and solving for c:
c = -1/2
after putting the value of c back in and rearranging i get:
y = √x^2 + x + 1/2
clearly i am doing something wrong. please help show how to get the correct answer!!!
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Hey hi friend,
I think this is quite simple question you asked. I will explain it to you.
dy/dx= (y+1)/(x+1)
So in variable separable we separate dy on one side and dx on one side.
Now we have,
1/(y+1)dy=1/(x+1)dx
taking log both side we get,
ln|y+1|=ln|x+1|+C (read integration formula)
now move ln|x+1| to other side and keep constant at other side.
ln|y+1|-ln|x+1=C
or, ln|(y+1)/(x+1)|=C
|(y+1)/(x+1)|=e to the power C (where e is exponential, when log is moved to other side it become like this)
Now whenGreatest integer fraction is removed then it become
(y+1)/(x+1)=+-e to the power C which could be written as A.
(y+1)/(x+1)=A
Now put value of x and y then A=2
Put value of A,
y+1=2(x+1)
y+1=2x+2
y=1+2x.
Hence done
See it was quite easy
Feel free to contact me at http://www.indiastudychannel.com/r/45967… to know more thing. Also i will tell you how to earn from below site. You will find my facebook link on my profile at this site.
I think this is quite simple question you asked. I will explain it to you.
dy/dx= (y+1)/(x+1)
So in variable separable we separate dy on one side and dx on one side.
Now we have,
1/(y+1)dy=1/(x+1)dx
taking log both side we get,
ln|y+1|=ln|x+1|+C (read integration formula)
now move ln|x+1| to other side and keep constant at other side.
ln|y+1|-ln|x+1=C
or, ln|(y+1)/(x+1)|=C
|(y+1)/(x+1)|=e to the power C (where e is exponential, when log is moved to other side it become like this)
Now whenGreatest integer fraction is removed then it become
(y+1)/(x+1)=+-e to the power C which could be written as A.
(y+1)/(x+1)=A
Now put value of x and y then A=2
Put value of A,
y+1=2(x+1)
y+1=2x+2
y=1+2x.
Hence done
See it was quite easy
Feel free to contact me at http://www.indiastudychannel.com/r/45967… to know more thing. Also i will tell you how to earn from below site. You will find my facebook link on my profile at this site.
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y[x] = 1 + 2 x