## anonymous 3 years ago Solve dy/dx = (x + 4y -1)^2

1. anonymous

solve $$\frac{ dy }{ dx } = (x + 4y -1)^{2}$$

2. carson889

It is 2x + 8y - 2.

3. carson889

Got it by expanding the brackets: x^2 + 4yx - x + 4xy + 16y^2 -4y -x - 4y + 1. Which simplifies to: x^2 + 8xy + 16y^2 - 2x -8y + 1. Taking the derivative of that with respect to x: 2x + 8y + 0 - 2 - 0 + 0 Which simplifies to 2x + 8y -2 or 2(x+4y-1)

4. carson889

Or using chain rule: $2(x+4y-1)^{1}(\frac{ d }{ dx }(x+4y-1))$ $\frac{ d }{ dx } = 1$ Therfore, $2(x+4y-1)(1) = 2x + 8y - 2$

5. anonymous

2(x+4y-1)=2x+8y-2.

6. anonymous

ok thank all a lots i appreciate it :)

7. anonymous

No Problem.

8. anonymous

@carson889: wait i mean $\frac{ dy }{ dx } = (x+4y - 1)^{}$ $$(x+4y -1)^{2}$$ is the result of $$\frac{ dy }{ dx }$$ it isn't $\frac{ d }{ dx} (x +4y -1)^{2}$ any idea???

9. anonymous

You need to use the chain rule, as this is implicit differentiation, not partial.

10. anonymous

what did u get it??

11. anonymous

So, are we trying to find $$y$$ ?

12. anonymous

(As a function of $$x$$ )

13. anonymous
14. anonymous

Yeah, so it's an actually differential equation which you have to solve. Here:

15. anonymous

Define (all in terms of x): $v=4y+x\implies v'=4y'+1$So: $\frac{v'-1}{4}=(v-1)^2\implies\\ \frac{v'}{4v^2-8v+5}=1$Integrating both sides: $\int \frac{v'}{4v^2-8v+5}\;dx=\int 1\;dx=\\ \frac{\tan^{-1}(2v-2)}{2}=x+C$Solving for $$v$$: $v=\frac{\tan(2(x+C))+2}{2}$Substituting $$y$$: $y=\frac{\tan(2(x+C))-2x+2}{8}$ That took forever.

16. anonymous

awesome..., thank u @LolWolf

17. anonymous

Sure thing

18. anonymous

@LolWolf

19. anonymous

i tried to calculate on Maple 13, but, look at the picture below

20. anonymous

is not equal to (x+4y-1)^2

21. anonymous

This is the equation I get, when solving the differential with Mathematica:

22. anonymous

In real form:

23. anonymous

ok...,

24. anonymous

it is right, isn't it?