Homogeneous Differential Equation: First Order & Second Order

Ans. Φ = tan^-1\(\frac{y}{x}\) – sin²\(\frac{y}{x}\) or \(\frac{dy}{dx}\)= tanΦ = \(\frac{y}{x}\) – sin² \(\frac{y}{x}\)…….(1)

Since, this equation illustrates a homogeneous type of differential equation, by substituting y=vx in equation (1), we have

v + x \(\frac{dv}{dx}\) = v - sin² v

x\(\frac{dv}{dx}\) = – sin²v

\(\frac{dx}{x}\) = – cosec²vdv

By integrating both sides with respect to x and v, we have

\(\int \frac{dx}{x}\) =\(\int \) – cosec² vdv

ιnx = \(\frac{1}{tanv}\) + C…….(2)

When the curve passes through given point i.e., \((2, \frac{\pi} {3})\), for (x,y). We know, v =\(\frac{y}{x}\), thus 

v = \(\frac{\pi}{3}\) ÷ 2 = \(\frac{\pi}{6}\)

Putting the values of v and x in equation (2), we have

ln2 = \(\sqrt{3}\) + C

Or lnx = \(\frac{1}{tanv}\) + ln2 – \(\sqrt{3}\) or lnx = \(\frac{1}{tan \frac{y}{x}}\) + ln2 – \(\sqrt{3}\) which is the solution for given equation. 

Ans. The given equation is \(x Cos (\frac{y}{x}) . \frac{dy}{dx} = yCos (\frac{y}{x}) + x\)

Here by separating the variables on either side of the equal-to symbol, we get

x . \(\frac{dy}{dx}\) = \(\frac{1}{Cosv}\)

Cos v.dv = \(\frac{dx}{x}\)

Integrating this expression on both sides, we get 

\(\int \)Cosv.dv = \(\int \)1x.dx \(\int \)Cosv.dv = 1x.dx

Siny = Logx + C

Again, putting y/x = v.

Sin\(\frac{y}{x}\) = Logx + C

Thus, the solution of the given equation is Sin\(\frac{y}{x}\) = Logx + C.

Ans. x dy= (x + y) dx…….(1)

By rearranging the given equation, we get

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

\(\frac{dy}{dx}\)= 1 + \(\frac{y}{x}\)

\(\frac{dy}{dx}\) = f (\(\frac{y}{x}\))

The differential equation given is a homogeneous equation, substituting y = vx, we get 

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

v + x \(\frac{dv}{dx}\)  = 1 + \(\frac{vx}{x}\)

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

x\(\frac{dv}{dx}\) = 1

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

By integrating both sides we get, 

\(\int \)dv =\(\int \) \(\frac{dx}{x}\) = C

v = ln \(\mid\)x\(\mid\) + C

Substituting v =\(\frac{y}{x}\), we get

\(\frac{y}{x}\)= In \(\mid\)x\(\mid\) + C

y = x In\(\mid\)x\(\mid\) + cx

∴ We have got the solution for given equation. 

Ans. By re-arranging the given equation, we get 

\(\frac{dy}{dx}\)= \(\frac{x^2 +3xy+y^2}{ x^2}\)

\(\frac{dy}{dx}\) = f (\(\frac{y}{x}\))

As the given equation is homogeneous equation, substitute y=vx, we have

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

v + x\(\frac{dv}{dx}\)= 1 + 3 \(\frac{vx}{x}\) + x \(\frac{(vx)^2}{x^2}\)

x\(\frac{dv}{dx}\) = 1 + 3v + v² = 1 + 2v + v²

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

= \(\frac{dx}{x}\)

By integrating both sides, we have

\(\int \frac{dv} {1+ 2v + v^2}\) = \(\frac{dx}{x}\) + c'

\(\int \frac{dv}{(v+1)^2}\) = \(\frac{dx}{x}\) + c'

\(\frac{x}{y+x}\) + ln\(\mid\) x \(\mid\)= c, is the solution for given homogeneous equation.

Ans. We can rewrite the given equation as \(\frac{dy}{dx}\)= \(\frac{x+2y}{x-y}\) …….(1)

Thus, we have solved the given equation and got the general solution for the same. 

Ans. The given equation can be written as 

Now, cancel v on both sides, we have

y + \(\sqrt{1 + v^2}\) = C₁x², which is the solution after substituting the value of v = y/x.

Ans. In a homogeneous differential equation there is a homogeneous function, say f(x,y), such that 

f (λx,λy) = λ ⁿ f (x,y), for any non-zero constant λ. Whereas, in a non-homogeneous differential equation, there is no homogeneous function. Example, linear differential equation of the form \(\frac{dy}{dx}\) + Py = Q is a non-homogeneous differential equation. 

Ans. We can re-write the given equation as:

Thus we have obtained the required solution for the given equation after substituting the value of v = y/x.

Ans. As each of the functions (y² – x²) and 2xy is a homogeneous function of degree 2, the given equation is homogeneous.

Substituting y = vx and dy/dx = v + x dy/dx, we get

Putting the value of v=y/x, we have

x (\(\frac{1+y^2}{x^2}\)) = C1

(x² + y²)= x C1, is the solution for the given equation. 

Ans. We can re-write the given equation as,

By taking ± C = C₁, we get

x² v cos cos v = C₁

xy cos( \(\frac{y}{x}\)) = C₁

Thus, we have got the required solution after substituting the actual value of v as y/x. 

Ans. The different types of differential equation are as follows:

• Ordinary differential equations 
• Partial differential equations 
• Linear differential equations 
• Nonlinear differential equations 
• Homogeneous differential equations 
• Nonhomogeneous differential equations 

Ans. The general equation of Homogeneous Differential Equation is as follows:

f(x, y).dy + g(x, y).dx = 0

The general equation of Non-Homogeneous Differential Equation is as follows:

y”+p(t)y’+q(t)y = g(t)