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Differential Equations

Long Answer Type

201. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

1 over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals 0

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202. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

2 space straight y space straight e to the power of straight x over 4 end exponent space dx plus open parentheses straight y minus 2 space straight x space straight e to the power of straight x over straight y end exponent close parentheses space dy space equals space 0

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203. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:

straight y space dx space plus space straight x space open parentheses log space straight y over straight x close parentheses space dy space minus space 2 space straight x space dy space equals space 0

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204. Show that the given differential equation is homogeneous and solve it:

left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx

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Short Answer Type

205. Show that the given differential equation is homogeneous and solve it:

straight y apostrophe space equals space fraction numerator straight x plus straight y over denominator straight x end fraction

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206. Show that the given differential equation is homogeneous and solve it:
(x-y) dy - (x+y) dx = 0

The given differential equation is
(x-y) dy - (x+y) dx = 0 or (x-y) dy = (x+y) dy
or

dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction

Replacing x by

λx

and y by

λx comma space we space get comma

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator λx plus λy over denominator λx minus λy end fraction space equals space fraction numerator straight lambda left parenthesis straight x plus straight y right parenthesis over denominator straight lambda left parenthesis straight x minus straight y right parenthesis end fraction space equals space straight lambda degree space space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket

∴ F(x, y) is a homogeneous function of degree zero.
∴ given differential equation is a homogeneous differential equation.
Put y = v x so that $dy over dx space equals space straight v plus straight x dv over dx$
$therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight v plus straight x dv over dx space equals space fraction numerator straight x plus vx over denominator straight x minus vx end fraction space space or space space space straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction minus straight v therefore space space space straight x dv over dx space equals fraction numerator 1 plus straight v minus straight v plus straight v squared over denominator 1 minus straight v end fraction space space or space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 1 minus straight v end fraction$
Separating the variables,    $fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space 1 over straight x dx$

Integrating , $integral fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx$

or    $integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv space minus space 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx space space or space space tan to the power of negative 1 end exponent straight v minus 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space log space open vertical bar straight x close vertical bar plus straight c$

or $tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c$

or    $tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c space is space the space required space solution. space$