1. ## Differential Equations questions

(a) Verify that all members of the family $\displaystyle y = (5)^{1/2} (c - x^2)^{-1/2}$ are solutions of the differential equation.

(b) Find a solution of the initial-value problem.

$\displaystyle \frac {dy}{dx} = \frac {xy^3}{5} , y(0) = 4$

I do not know how to solve a. However, This is how I solved b. Please give me a hint to solve a and check my answer for b.

(b)

$\displaystyle \frac {dy}{dx} = \frac {xy^3}{5}$

Move the ys to one side:

$\displaystyle \frac {dy}{y^3} = \frac{xdx}{5}$

Integrate:

$\displaystyle \int \frac {dy}{y^3} = \int \frac{xdx}{5}$

$\displaystyle \frac {-2}{y^2} = \frac {x^2}{10} + C$

$\displaystyle y = \sqrt{\frac{-5}{x^2}} + C$

I dont think my answer is correct.

2. It's all right till to...

$\displaystyle \int \frac{dy}{y^{3}} = \int \frac{x}{5} \cdot dx$

Now you have to remember the integration rule of...

$\displaystyle \int y^{n}\cdot dy$

... and the fact that an indefinite integral contains an arbitrary constant usually indicated with $\displaystyle c$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Yes, I forgot the C. However, to integrate: $\displaystyle y^n$ isn't it a simple:

$\displaystyle \frac{y^{n+1}}{n+1}$

PS. How do you raise y to a n+1 with latex?

so finding c using y = 4 at x = 0
finally we get
y(x) = sqrt ( 80 / (5-4x^2) )

5. Originally Posted by calc101
Yes, I forgot the C. However, to integrate: $\displaystyle y^n$ isn't it a simple:

$\displaystyle \frac{y^ny^1}{n+1}$

PS. How do you raise y to a n+1 with latex?
Use { } around the n+1:
$\displaystyle \frac{y^{n+1}}{n+1}$
Click on that to see the code.

6. Originally Posted by calc101
(a) Verify that all members of the family $\displaystyle y = (5)^(1/2) (c - x2)^(-1/2)$ are solutions of the differential equation.
A solution to what differential equation? To the equation in (b)?
If $\displaystyle y= \sqrt{5}(c- x^2)^{-\frac{1}{2}}$ then $\displaystyle \frac{dy}{dx}= \sqrt{5}(-\frac{1}{2})(c- x^2)^{-3/2}(-2x)$ while $\displaystyle y^3= 5\sqrt{5}(c- x^2)^{-\frac{3}{2}}$ so $\displaystyle \frac{xy^3}{5}= \sqrt{5}x(c-x^2)^{-\frac{3}{2}}$.

(b) Find a solution of the initial-value problem.

$\displaystyle \frac {dy}{dx} = \frac {xy^3}{5} , y(0) = 4$

I do not know how to solve a. However, This is how I solved b. Please give me a hint to solve a and check my answer for b.

(b)

$\displaystyle \frac {dy}{dx} = \frac {xy^3}{5}$

Move the ys to one side:

$\displaystyle \frac {dy}{y^3} = \frac{xdx}{5}$

Integrate:

$\displaystyle \int \frac {dy}{y^3} = \int \frac{xdx}{5}$

$\displaystyle \frac {-2}{y^2} = \frac {x^2}{10} + C$

$\displaystyle y = \sqrt{\frac{-5}{x^2}}$

I dont think my answer is correct.

7. Originally Posted by vinay
so finding c using y = 4 at x = 0
finally we get
y(x) = 20 / (5-x^2)

Thanks,

8. Move the ys to one side:

Integrate:

--------------------- eq 1

so put x = 0, and y(0) = 4

which gives you value of c

c = -2/4^2 = -1/8

now again subsitute c in main eq 1

-2/y^2 = x^2/10 -1/8

which will give the final answer as,

y(x) = sqrt ( 80 / (5-4x^2) )

-

9. Thanks Vinay

$\displaystyle y(x) = \sqrt { \frac {80} {5-4x^2} }$