# Differential equation

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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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#2

(Original post by

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

**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

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(Original post by

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.

**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.

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#4

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I did that i got -1/y=x^2/2+c. I don't know how to go ahead

**Shas72**)I did that i got -1/y=x^2/2+c. I don't know how to go ahead

Upload your working if you want it checked.

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#5

(Original post by

I did that i got -1/y=x^2/2+c. I don't know how to go ahead

**Shas72**)I did that i got -1/y=x^2/2+c. I don't know how to go ahead

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(Original post by

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

**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

Assuming its right, use the point (1,3) to get c.

Upload your working if you want it checked.

**mqb2766**)Assuming its right, use the point (1,3) to get c.

Upload your working if you want it checked.

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#7

(Original post by

I got c=-5/6

**Shas72**)I got c=-5/6

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(Original post by

Upload your calcs if you want them checked.

**mqb2766**)Upload your calcs if you want them checked.

1/y^2.dy=x.dx

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

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#9

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(Original post by

You've not used the proportional to information with the fact that the graident is 6 at (1,3)

**mqb2766**)You've not used the proportional to information with the fact that the graident is 6 at (1,3)

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#11

(Original post by

I did not understand. Do I just say proportional or equal to and solve

**Shas72**)I did not understand. Do I just say proportional or equal to and solve

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(Original post by

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?

**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?

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**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?

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#14

(Original post by

So when I do that I get c= 2/3

**Shas72**)So when I do that I get c= 2/3

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(Original post by

Again, upload what you've done if you want it checked.

**mqb2766**)Again, upload what you've done if you want it checked.

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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#16

(Original post by

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

**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

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(Original post by

Looks ok.

**mqb2766**)Looks ok.

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#18

(Original post by

Thanks. Iam stuck with one more problem. Pls help.its the 14th sum

**Shas72**)Thanks. Iam stuck with one more problem. Pls help.its the 14th sum

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(Original post by

Have you started by separating the variables? Post where you're up to?

**mqb2766**)Have you started by separating the variables? Post where you're up to?

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