Math, asked by anjani169, 8 months ago

solution of the differential equation x dy/dx = y ln(y^2/x^2) is​

Answers

Answered by anushkaganguly477
2

e^(x u(x)) x (x (( du(x))/( dx) - 2) + u(x)) = e^(x u(x)) x u(x)

Solve for ( du(x))/( dx):

( du(x))/( dx) = 2

Integrate both sides with respect to x:

u(x) = integral2 dx = 2 x + c_1, where c_1 is an arbitrary constant.

Substitute back for v(x) = x u(x):

v(x) = x (2 x + c_1)

Substitute back for y(x) = e^v(x), which gives v(x) = log(y(x)):

log(y(x)) = x (2 x + c_1)

Solve for y(x):

Answer: y(x) = e^(2 x^2 + c_1 x)

Answered by rinayjainsl
0

Answer:

The solution of differential equation is

y = xe {}^{ \frac{cx {}^{2} + 1 }{2} }

Step-by-step explanation:

Given that,the differential equation is

x \frac{dy}{dx}  = yln( \frac{y {}^{2} }{x {}^{2} } )

It can be reframed in the given way

x \frac{dy}{dx}  = 2yln( \frac{y}{x}) \\  =  >  \frac{dy}{dx}  = 2 \frac{y}{x} ln( \frac{y}{x} )

We use homogenous method to solve this differential equation by substituting

v =  \frac{y}{x}  =  > y = vx \\  =  >  \frac{dy}{dx}  = v + x \frac{dv}{dx}

Substituting these in the given differential equation,we get

v + x \frac{dv}{dx}  = 2vln(v) \\  =  >  \frac{dv}{v(2lnv - 1)}  =  \frac{dx}{x}

again

put \ \: : 2lnv - 1 = t \\  =  >  \frac{2}{v} dv = dt =  >  \frac{dv}{v}  =  \frac{1}{2} dt

Re substituting again,

 \frac{dt}{2t}  =  \frac{dx}{x}

Integrating on both sides,we get

\int \frac{dt}{2t}  = \int \frac{dx}{x}  \\  =  >  \frac{1}{2} lnt = lnx + c \\  =  > t = cx {}^{2}

Since t is known

2lnv - 1 = cx {}^{2}  \\  =  > 2ln \frac{y}{x}  - 1 = cx {}^{2}  \\  =  > ln \frac{y}{x}  =  \frac{cx {}^{2} + 1 }{2}  \\  =  > y = xe {}^{ \frac{cx {}^{2} + 1 }{2} }

Therefore,the solution of differential equation is

y = xe {}^{ \frac{cx {}^{2} + 1 }{2} }

#SPJ2

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