## Active forum topics

- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
- Application of Differential Equation: Newton's Law of Cooling
- Minima maxima: a²y = x⁴
- Trim and stability

## New forum topics

- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
- Required diameter of solid shaft
- Differentiate trigonometric function
- Integration $\displaystyle \int \sec 2x ~ dx$

## Recent comments

- Mali po ang equation mo…3 days 9 hours ago
- $x$ is the location where…3 days 20 hours ago
- In double integration method…1 week 1 day ago
- Maayo, salamat sa imong…2 weeks 3 days ago
- 24 ft during the 10th second…2 weeks 3 days ago
- The differentiation you need…2 weeks 4 days ago
- Obtain the differential…2 weeks 4 days ago
- Thank you for sharing your…2 weeks 4 days ago
- Based on the differentiation…1 week 1 day ago
- Given that $x + y + xy = 1$,…1 week 1 day ago

## Re: Differential Eqn.

Let r = radius of cirlce

Equation of family of circles of radius r and tangent to the y-axis:

(x ± r)

^{2}+ (y - k)^{2}= r^{2}The (±) sign indicates that circles can be at the left or at the right of y-axis. There is only one arbitrary constant k, thus, the differential equation is a first degree. Note that r here is a parameter (fixed radius) and need not be eliminated.

2(x ± r) + 2(y - k)y' = 0

(x ± r) + (y - k)y' = 0

(y - k) = -(x ± r) / y'

Thus,

$(x \pm r)^2 + \left( -\dfrac{x \pm r}{y'} \right)^2 = r^2$

$(x \pm r)^2 + \dfrac{(x \pm r)^2}{(y')^2} = r^2$

$(x \pm r)^2 (y')^2 + (x \pm r)^2 = r^2 (y')^2$

$r^2 (y')^2 - (x \pm r)^2 (y')^2 = (x \pm r)^2$

$[ \, r^2 - (x \pm r)^2 \, ] (y')^2 = (x \pm r)^2$

$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{(x \pm r)^2}{r^2 - (x \pm r)^2}$

$\dfrac{dy}{dx} = \dfrac{x \pm r}{\sqrt{r^2 - (x \pm r)^2}}$

answerPlease double check my answer. I am answering here out from what I have remembered from this topic. It's been a long while since I encountered this type of problem.