Homogeneous Differential Equations

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What is homogeneous differential equation?

If a first order, first degree differential equation is expressible in the form

\dpi{120} \mathbf{\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}}

where f(x,y) and g(x,y) are homogeneous functions of the same degree, then  it is called a homogeneous differential equation.

Such type of equations can be reduced to variable separable form by some suitable substitution  like y=vx.

What is homogeneous function?

A function f(x,y) is called a homogeneous function of degree n if

\dpi{120} \mathbf{f(\lambda x,\lambda y)=\lambda ^{n}f(x,y)}

For example ,

\dpi{120} f(x,y)=x^{2}-y^{2}+5xy

is a homogeneous function of degree 2 because

\dpi{120} f(\lambda x,\lambda y)=\lambda ^{2}x^{2}-\lambda ^{2}y^{2}+5\lambda x\lambda y=\lambda ^{2}f(x,y)

 

Example1: Solve the differential equation    (x+y)dy+(x-y)dx=0  given that y=1 when x=1.

Solution:  (x+y)dy+(x-y)dx=0

\frac{dy}{dx}=\frac{y-x}{x+y}

Here both numerator y-x and denominator x+y  are functions of degree 1 , therefore this is homogeneous differential equation.

Using substitution,  y=vx

\frac{dy}{dx}=v+x\frac{dv}{dx}

Original  DE get changed  to,

v+x\frac{dv}{dx}=\frac{vx-x}{x+vx}

Dividing numerator and denominator with x and subtracting v we get,

x\frac{dv}{dx}=\frac{v-1}{1+v}-v

x\frac{dv}{dx}=\frac{v-1-v^{2}-v}{1+v}

x\frac{dv}{dx}=\frac{-(v^{2}+1)}{1+v}

\frac{v+1}{v^{2}+1}dv =\frac{-dx}{x}

\int \frac{v+1}{v^{2}+1}dv =\int \frac{-dx}{x}

\int \frac{v}{v^{2}+1}dv +\int \frac{1}{v^{2}+1}dv=-\int \frac{dx}{x}

\frac{1}{2}\int \frac{2v}{v^{2}+1}dv +\int \frac{1}{v^{2}+1}dv=-\int \frac{dx}{x}

\frac{1}{2}ln(v^{2}+1)+tan^{-1}(v)=-ln|x|+C

ln(v^{2}+1)+2ln|x|+2tan^{-1}(v)=2C

ln(v^{2}+1)+lnx^{2}+2tan^{-1}(v)=2C

ln[(v^{2}+1)x^{2}]+2tan^{-1}(v)=2C

Plugin back v as y/x, we get expression as

ln[(\frac{y^{2}}{x^{2}}+1)x^{2}]+2tan^{-1}(\frac{y}{x})= K where K=2C

ln(y^{2}+x^{2})+2tan^{-1}(\frac{y}{x})= K

Using given condition that y=1 when x=1 we get,

ln(1+1)+2tan^{-1}(1)= K

ln(2)+2(\frac{\pi }{4})= K

ln(2)+\frac{\pi }{2}= K

Using value of constant K we get the solution as,

ln(y^{2}+x^{2})+2tan^{-1}(\frac{y}{x})= ln(2)+\frac{\pi }{2}

 

Example2: Solve the following differential equation.

{\color{Red} (4 x^{4}+x^{3}y+y^{4})dx-x^{4}dy=0}

Solution:

(4 x^{4}+x^{3}y+y^{4})dx=x^{4}dy

(4 x^{4}+x^{3}y+y^{4})dx=x^{4}dy

\frac{dy}{dx}=\frac{(4 x^{4}+x^{3}y+y^{4})}{x^{4}}

Substitute y=vx   and we get the above equation transformed as,

v+x\frac{dv}{dx}=\frac{(4 x^{4}+x^{4}v+v^{4}x^{4})}{x^{4}}

v+x\frac{dv}{dx}= 4+v+v^{4}

x\frac{dv}{dx}= 4+v^{4}

\frac{dv}{4+v^{4}}= \frac{dx}{x}

\int \frac{dv}{4+v^{4}}= \int \frac{dx}{x}

\int \frac{dv}{2^{2}+(v^{2})^{2}}= \int \frac{dx}{x}

\frac{1}{2}tan^{-1}\left ( \frac{v^{2}}{2} \right )=ln|x|+C

Plugin back v=y/x

\frac{1}{2}tan^{-1}\left ( \frac{y^{2}}{2x^{2}} \right )=ln|x|+C

 

Example3: Solve the following homogeneous differential equation

{\color{Red} xsin\left ( \frac{y}{x} \right )\frac{dy}{dx}=ysin\left ( \frac{y}{x} \right )+x}

Solution:

\frac{dy}{dx}=\frac{ysin\left ( \frac{y}{x} \right )+x}{ x sin\left ( \frac{y}{x} \right )}

Substitute y=vx  and we get the above equation transformed as,

v+x\frac{dv}{dx}=\frac{(vx) sin\left ( \frac{vx}{x} \right )+x}{ x sin\left ( \frac{vx}{x} \right )}

v+x\frac{dv}{dx}=\frac{(vx) sin\left ( v\right )+x}{ x sin\left (v \right )}

v+x\frac{dv}{dx}=\frac{ x(v sin\left ( v\right )+1)}{ x sin\left (v \right )}

v+x\frac{dv}{dx}=\frac{ v sin\left ( v\right )+1}{ sin\left (v \right )}

x\frac{dv}{dx}=\frac{ v sin\left ( v\right )+1}{ sin\left (v \right )}-v

x\frac{dv}{dx}=\frac{ v sin\left ( v\right )+1-vsin(v)}{ sin\left (v \right )}

x\frac{dv}{dx}=\frac{ 1}{ sin\left (v \right )}

sin(v)dv= \frac{dx}{x}

\int sin(v)dv= \int \frac{dx}{x}

-cos(v) = ln|x|+C

cos(y/x)=-ln|x|-C

 

Sometimes a homogeneous differential equation is  expressible in the form,

\dpi{120} \frac{dx}{dy}=\frac{f(x,y)}{g(x,y)}

In such a situation, we substitute x=vy  and proceed same way .

 

Example4: Solve

\dpi{120} {\color{Red} 2ye^{\frac{x}{y}}dx +(y-2xe^{\frac{x}{y}})dy=0}

Solution:

\frac{dx}{dy}=\frac{2xe\frac{x}{y}-y}{2ye^{\frac{x}{y}}}

Let x=vy, then

\frac{dx}{dy}=v+y\frac{dv}{dy}

And the given DE can be rewritten as,

v+y\frac{dv}{dy}=\frac{2vye^{v}-y}{2ye^{v}}

y\frac{dv}{dy}=\frac{2ve^{v}-1}{2e^{v}}-v

y\frac{dv}{dy}=\frac{2ve^{v}-1-2ve^{v}}{2e^{v}}

y\frac{dv}{dy}=\frac{-1}{2e^{v}}

2e^{v}dv=-\frac{1}{y}dy

\int 2e^{v}dv=-\int \frac{1}{y}dy

2e^{v}=- \ln (y)+lnC

2e^{\frac{x}{y}}=ln|\frac{c}{y}|

 

 

 

 

Practice problems:

Solve the following homogeneous differential equations.

1. 2xy\frac{dy}{dx}=x^{2}+y^{2}

2.  xyln\left (\frac{x}{y} \right )dx+\left ( y^{2}-x^{2} ln\left ( \frac{x}{y} \right )\right )dy=0

3.  xdy-ydx=\sqrt{x^{2}+y^{2}}dx

Solve the following initial value problem

4. x(x^{2}+3y^{2})dx+y(y^{2}+3x^{2})dy=0 ;y(1)=1

 

 

 

 

 

 

Answers:

1. x=c(x^{2}-y^{2})

2. \frac{x^{2}}{y^{2}}\left ( ln\left ( \frac{x}{y}\right )-\frac{1}{2} \right )+lny^{2}=c

3. {y+\sqrt{x^{2}+y^{2}}}= cx^{2}

4. x^{4}+6x^{2}y^{2}+y^{4}=8

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