A function of the form \( F(x,y) \)

Or

A differential equation of the form \( \frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}\)

Consider the following functions in x and y,

\(F_{1} (x,y) = 2x- 8y \)

\(F_{2} (x,y) = x^{2}+ 8xy +9y^{2}\)

\(F_{3} (x,y) = \sin \left ( \frac{x}{y} \right )\)

\(F_{4} (x,y) = \sin x + \cos y\)

If we replace x and y with vx and vy respectively, for non-zero value of v, we get

\(F_{1} (vx,vy) = 2(vx)- 8(vy) = v(2x- 8y) =v \; F_{1} (x,y) \)

\(F_{2} (vx,vy) = v^{2}x^{2}+ 8(vx)(vy) +9v^{2}y^{2}=v^{2}(x^{2}+ 8xy +9y^{2}) = v^{2} F_{2} (x,y) \)

\(F_{3} (vx,vy) = \sin \left ( \frac{vx}{vy} \right ) = v^{0}\sin \left ( \frac{vx}{vy} \right )= v^{0}F_{3} (x,y) \)

\(F_{4} (vx,vy) = \sin (vx) + \cos (vy) \neq v^{n}F_{4}(x,y)\)

Hence, functions \(F_{1},F_{2},F_{3}\)

Solving Homogeneous Differential Equation-

To solve a homogeneous differential equation following steps are followed:-

Given differential equation of the type \(\frac{\mathrm{d} y}{\mathrm{d} x} = F(x,y) = g\left ( \frac{y}{x} \right )\)

Step 1- Substitute \( y = vx \)

Step 2 – Differentiating, we get, \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

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

\(\Rightarrow x\frac{dv}{dx} = g(v)- v\)

Step 3 – Separating the variables, we get

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

Step 4 – Integrating both side of equation, we have

\(\int \frac{dv}{g(v)-v} dv =\int \frac{dx}{x} + C\)

Step 5 – After integration we replace \(v = \frac{y}{x}\)

Lets Work Out- Example: Find the equation of the curve passing through the point \((2, \frac{\pi}{3} )\) Solution: \(\phi = tan^-1 (\frac{y}{x} – sin^2 \frac{y}{x})\) Or \(\frac{dy}{dx} = tan\phi = \frac{y}{x} – sin^2 \frac{y}{x} \) Since this equation represents a differential equation of homogeneous type therefore we substitute \( y = vx \) \(\Rightarrow v + x\frac{dv}{dx} = v – sin^2 v\) \(\Rightarrow x \frac{dv}{dx} = – sin^2 v\) \(\Rightarrow \frac{dx}{x} = -cosec^2 v dv \) Now integrating both the sides w.r.t. to x and v respectively, we get \( \int \frac{dx}{x} = \int -cosec^2 v dv \) \( ln x = \frac{1}{tan v} + C \) Also as it passes through the point \( (2, \frac{\pi}{3})\) We know that v = y/x, thus value of v = \(\frac{\pi}{3} \div 2 = \frac{\pi}{6}\) So substituting the values of x and v in the equation (i), we get \( ln 2 = \sqrt{3} + C \) \( \Rightarrow C = ln 2 – \sqrt{3} \) Or \( ln x = \frac{1}{tan v} + ln 2 – \sqrt{3} \) Or \( ln x = \frac{1}{tan\frac{y}{x}} + ln 2 – \sqrt{3} \) This is the required solution. Example: Find the equation of the curve passing through the point (1,-2) when the tangent at any point is given by \( \frac{y(x + y^3)}{x(y^3 – x)} \) Solution: The equation of tangent represents the slope of the curve i.e. This equation is homogeneous in nature. On cross-multiplication, we get- \( (xy^3 – x^2)dy = (xy + y^4)dx \) Solving the equation, we get \(x^{2}y^{3} \frac{\left( xdy – ydx \right )}{x^{2}} – x (xdy – ydx) = 0\) \(\Rightarrow x^{2}y^{3} d\frac{y}{x} – xd(xy) = 0\) Dividing both the sides by \( x^3y^2 \) \( \frac{y}{x} d(\frac{y}{x}) – \frac{d(xy)}{x^2y^2} = 0 \) Now integrating this equation with respect to \(\frac{y}{x} and xy \) \( \int \frac{y}{x} d(\frac{y}{x}) = \int \frac{d(xy)}{x^2y^2}\) \(\frac{1}{2} \left ( \frac{y}{x} \right )^{2} = – \frac{1}{xy} +C\) Now substituting the value of the given point in the above equation, we have \( \Rightarrow \frac{1}{2} \times 4 – \frac{1}{2} = C \) \( \Rightarrow C = \frac{3}{2} \) Put this value of the constant C in equation (1) we get \( \frac{1}{2} (\frac{y}{x})^2 + \frac{1}{xy} = \frac{3}{2} \) This is the required solution. |

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