# Differential equation

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#1
at a point with coordinates (x,y), the gradient of the curve is directly propotional to xy^2. At a particular point P with coordinates (1,3) its known that the gradient of this curve is 6. find and simplify an expression for y in terms of x
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3 weeks ago
#2
(Original post by Shas72)
at a point with coordinates (x,y), the gradient of the curve is directly propotional to xy^2. At a particular point P with coordinates (1,3) its known that the gradient of this curve is 6. find and simplify an expression for y in terms of x
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and the proportionality constant.
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#3
(Original post by mqb2766)
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and the proportionality constant.
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
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3 weeks ago
#4
(Original post by Shas72)
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
Assuming its right, use the point (1,3) to get c.
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3 weeks ago
#5
(Original post by Shas72)
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
Rearrange to make y the subject, and using the initial conditions given that dy/dx = 6 at (1,3), solve for c
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#6
(Original post by mqb2766)
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and
(Original post by mqb2766)
Assuming its right, use the point (1,3) to get c.
I got c=-5/6
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3 weeks ago
#7
(Original post by Shas72)
I got c=-5/6
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#8
(Original post by mqb2766)
Ok I did dy/dx=xy^2
1/y^2.dy=x.dx
-1/y=x^2/2+c
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3 weeks ago
#9
(Original post by Shas72)
Ok I did dy/dx=xy^2
1/y^2.dy=x.dx
-1/y=x^2/2+c
You've not used the proportional to information with the fact that the graident is 6 at (1,3)
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#10
(Original post by mqb2766)
You've not used the proportional to information with the fact that the graident is 6 at (1,3)
I did not understand. Do I just say proportional or equal to and solve
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3 weeks ago
#11
(Original post by Shas72)
I did not understand. Do I just say proportional or equal to and solve
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
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#12
(Original post by mqb2766)
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
Ok let me try. I seem to have forgotten. Let me try
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#13
(Original post by mqb2766)
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
So when I do that I get c= 2/3
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3 weeks ago
#14
(Original post by Shas72)
So when I do that I get c= 2/3
Again, upload what you've done if you want it checked.
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#15
(Original post by mqb2766)
Again, upload what you've done if you want it checked.
So I did dy/dx=k.xy^2
Substituting the values I got k=2/3
Then I did dy/dx=2/3× xy^2
I got-1/y=x^2/3+c
Substituting (1,3) I get c=-2/3
Hence-1/y=x^2/3-2/3
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3 weeks ago
#16
(Original post by Shas72)
So I did dy/dx=k.xy^2
Substituting the values I got k=2/3
Then I did dy/dx=2/3× xy^2
I got-1/y=x^2/3+c
Substituting (1,3) I get c=-2/3
Hence-1/y=x^2/3-2/3
Looks ok.
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#17
(Original post by mqb2766)
Looks ok.
Thanks. Iam stuck with one more problem. Pls help.its the 14th sum
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3 weeks ago
#18
(Original post by Shas72)
Thanks. Iam stuck with one more problem. Pls help.its the 14th sum
Have you started by separating the variables? Post where you're up to?
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#19
(Original post by mqb2766)
Have you started by separating the variables? Post where you're up to?
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#20
So when I do integration, I get lnx+1/2lnx^2+3=kt+c
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