Question

help plzzzz

Answer 1

i dunno

Answer 2

any ideas?

Answer 3

not sure if this will work out,
v = 4x+y+1
dv/dx = 4 + dy/dx => dy/dx = dv/dx - 4

=> dv/dx - 4 = v^2

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..

Not sure if this will work.
wolfram is giving different answer!!
http://www.wolframalpha.com/input/?i=y%27+%3D+%284x%2By%2B1%29%5E2

Answer 4

\[ \frac{dv}{4 + v^2} = dx\]
Integrating both sides,
\[ \tan^-1(\frac v 2) = x +c\implies 4x+y+1 = \tan(x+c)\]

Not sure if this will work ..!!

Answer 5

Let 4x+y+1=v
Then y=v-4x-1
and dy/dx=dv/dx-4
the DE becomes
\[\large{\frac{dy}{dx}=v^2+4}\]
which is a separable
\[\large{\frac{dv}{v^2+4} =dx}\]
Integrate both sides
\[\frac{1}{2}\tan^{-1}(\frac{v}{2})=x+C\]
\[v= 2\tan (2x+C)\]
therefore \[y=2\tan (2x+C) - 4x -1\]

Answer 6

looks like i missed 1/a part

Answer 7

yep
i din see the problem in the correct way
@experimentX and @lalaly thanks you for your help.

Answer 8

yw

Answer 9

:)