Solve the Following Initial Value Problem: (Xy − Y2) Dx − X2 Dy = 0, Y(1) = 1 - Mathematics

Question

Solve the following initial value problem:
(xy − y²) dx − x² dy = 0, y(1) = 1

Solution

(xy − y²) dx − x² dy = 0, y(1) = 1
This is an homogenous equation, put y = vx
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
$(x y - y^{2}) = x^{2} (\frac{d y}{d x})$
$(v x^{2} - v^{2} x^{2}) = x^{2} (v + x \frac{d v}{d x})$
$v x^{2} (1 - v) = x^{2} (v + x \frac{d v}{d x})$
$v (1 - v) = v + x \frac{d v}{d x}$
$v - v^{2} = v + x \frac{d v}{d x}$
$- v^{2} = x \frac{d v}{d x}$
$- \frac{1}{x} d x = \frac{1}{v^{2}} d v$
On integrating both sides we get,
$- \int \frac{1}{x} d x = \int \frac{1}{v^{2}} d v$
$- \log_{e} x = \frac{v^{- 2 + 1}}{- 2 + 1} + c$
$- \log_{e} x = \frac{v^{- 1}}{- 1} + c$
$- \log_{e} x = - \frac{1}{v} + c$
$- \log_{e} x = - \frac{1}{v} + c$
$\frac{x}{y} - \log_{e} x = c$
$As y (1) = 1$
$\frac{1}{1} - \log_{e} 1 = c$
$\Rightarrow c = 1$

shaalaa.com

Methods of Solving First Order, First Degree Differential Equations - Homogeneous Differential Equations

Is there an error in this question or solution?

Q 36.4 Q 36.3 Q 36.5

Chapter 22: Differential Equations - Exercise 22.09 [Page 84]

APPEARS IN

RD Sharma Mathematics [English] Class 12

Chapter 22 Differential Equations
Exercise 22.09 | Q 36.4 | Page 84

Video TutorialsVIEW ALL [3]

RELATED QUESTIONS

Solve the differential equation :

$y + x \frac{d y}{d x} = x - y \frac{d y}{d x}$

Show that the given differential equation is homogeneous and solve them.

$x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y$

Show that the given differential equation is homogeneous and solve them.

$x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = 0$

Show that the given differential equation is homogeneous and solve them.

$(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$

For the differential equation find a particular solution satisfying the given condition:

x² dy + (xy + y²) dx = 0; y = 1 when x = 1

For the differential equation find a particular solution satisfying the given condition:

$[x \sin^{2} (\frac{y}{x} - y)] d x + x d y = 0; y = \frac{π}{4} when x = 1$

For the differential equation find a particular solution satisfying the given condition:

$\frac{d y}{d x} - \frac{y}{x} + \cos e c (\frac{y}{x}) = 0; y = 0$ when x = 1

For the differential equation find a particular solution satisfying the given condition:

$2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0; y = 2$ when x = 1

Prove that x² – y² = c(x² + y²)² is the general solution of the differential equation (x³ – 3xy²)dx = (y³ – 3x²y)dy, where C is parameter

x y \log (\frac{x}{y}) d x + {y^{2} - x^{2} \log (\frac{x}{y})} d y = 0

x \cos (\frac{y}{x}) \cdot (y d x + x d y) = y \sin (\frac{y}{x}) \cdot (x d y - y d x)

(2x² y + y³) dx + (xy² − 3x³) dy = 0

Solve the following initial value problem:
(x² + y²) dx = 2xy dy, y (1) = 0

Solve the following initial value problem:
(y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1

Solve the following initial value problem:
x (x² + 3y²) dx + y (y² + 3x²) dy = 0, y (1) = 1

Find the particular solution of the differential equation x cos $(\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x$ , given that when x = 1, $y = \frac{π}{4}$

Find the particular solution of the differential equation $(x - y) \frac{d y}{d x} = x + 2 y$ , given that when x = 1, y = 0.

Show that the family of curves for which $\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$ , is given by $x^{2} - y^{2} = C x$

A homogeneous differential equation of the form $\frac{d x}{d y} = h (\frac{x}{y})$ can be solved by making the substitution

Solve the differential equation: x dy - y dx = $\sqrt{x^{2} + y^{2}} d x,$ given that y = 0 when x = 1.