## Active forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- 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

## New forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- 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

## Recent comments

- 400000=120[14π(D2−10000)]

(…1 week 5 days ago - Use integration by parts for…1 month 1 week ago
- need answer1 month 1 week ago
- Yes you are absolutely right…1 month 1 week ago
- I think what is ask is the…1 month 1 week ago
- $\cos \theta = \dfrac{2}{…1 month 1 week ago
- Why did you use (1/SQ root 5…1 month 1 week ago
- How did you get the 300 000pi1 month 1 week ago
- It is not necessary to…1 month 2 weeks ago
- Draw a horizontal time line…1 month 2 weeks ago

## What do you need to do with

What do you need to do with this equation? You just write the equation but there is no instruction.

## Can i solve this using homo?

In reply to What do you need to do with by Jhun Vert

Can i solve this using homo? Our professor only gave that equation and nothing else.

## Nope use suggested

In reply to Can i solve this using homo? by Helpme

Nope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y

## $$(x+2y-1)dx-(x+2y-5)dy=0$$

$$(x+2y-1)dx-(x+2y-5)dy=0$$

Let $u = x + 2y - 1 \rightarrow du = dx + 2dy$ which gives $dx = du - 2dy$

Substituting to the equation

$$\begin{eqnarray}

u(du - 2dy) - (u - 4)dy &=& 0\\

u du - (3u - 4)dy &=& 0\\

\dfrac{udu}{3u-4} - dy &=& 0\\

\dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)du - dy &=& 0\\

\displaystyle{\int \dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)}du - \int dy &=& C\\

\dfrac{1}{3}u - \dfrac{4}{3} \ln|3u - 4| - y &=& C\\

u - 4 \ln|3u - 4| - 3y &=& C\\

4 \ln|3u - 4| &=& u - 3y + C\\

4 \ln|3(x + 2y - 1) - 4| &=& x + 2y - 3y + C\\

4 \ln|3x + 6y - 7| &=& x - y + C\\

3x + 6y - 7 &=& Ce^{\frac{x-y}{4}} (ANSWER)

\end{eqnarray}$$

P.S. There can also be other forms of the general solution