Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[x \cos\left( \frac{y}{x} \right)\left( y dx + x dy \right) = y \sin \left( \frac{y}{x} \right)\left( x dy - y dx \right)\]
\[ \Rightarrow xy \cos \left( \frac{y}{x} \right) dx + x^2 \cos \left( \frac{y}{x} \right) dy = xy \sin \left( \frac{y}{x} \right) dy - y^2 \sin \left( \frac{y}{x} \right) dx\]
\[ \Rightarrow \left[ xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right) \right] dx = \left[ xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right) \right] dy\]
\[ \Rightarrow \frac{dy}{dx} = \frac{xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)}{xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{v x^2 \cos v + v^2 x^2 \sin v}{v x^2 \sin v - x^2 \cos v}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}\]
\[ \Rightarrow \frac{v\sin v - \cos v}{2 v \cos v}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{v\sin v - \cos v}{2 v \cos v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{v\sin v - \cos v}{v \cos v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{v \sin v}{v \cos v}dv - \int\frac{\cos v}{v \cos v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \int\tan v dv - \int\frac{1}{v}dv = 2\int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| \sec v \right| - \log \left| v \right| = 2 \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \frac{\sec v}{v} \right| = \log \left| C x^2 \right|\]
\[ \Rightarrow \frac{\sec v}{v} = C x^2 \]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[\sec \left( \frac{y}{x} \right) = \frac{y}{x} \times C \times x^2 \]
\[ \Rightarrow \sec \left( \frac{y}{x} \right) = Cxy\]
\[\text{Hence, }\sec \left( \frac{y}{x} \right) = Cxy\text{ is the required solution.}\]
APPEARS IN
संबंधित प्रश्न
Solve the differential equation (x2 + y2)dx- 2xydy = 0
Show that the differential equation 2yx/y dx + (y − 2x ex/y) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.
Show that the given differential equation is homogeneous and solve them.
`y' = (x + y)/x`
Show that the given differential equation is homogeneous and solve them.
(x – y) dy – (x + y) dx = 0
Show that the given differential equation is homogeneous and solve them.
`x dy - y dx = sqrt(x^2 + y^2) dx`
For the differential equation find a particular solution satisfying the given condition:
`[xsin^2(y/x - y)] dx + x dy = 0; y = pi/4 "when" x = 1`
Solve the following initial value problem:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]
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
Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]
A homogeneous differential equation of the form \[\frac{dx}{dy} = h\left( \frac{x}{y} \right)\] can be solved by making the substitution
Which of the following is a homogeneous differential equation?
Solve the differential equation: ` (dy)/(dx) = (x + y )/ (x - y )`
Solve the differential equation: x dy - y dx = `sqrt(x^2 + y^2)dx,` given that y = 0 when x = 1.
Solve the following differential equation:
y2 dx + (xy + x2)dy = 0
Solve the following differential equation:
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
Solve the following differential equation:
`x * dy/dx - y + x * sin(y/x) = 0`
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
State the type of the differential equation for the equation. xdy – ydx = `sqrt(x^2 + y^2) "d"x` and solve it
F(x, y) = `(x^2 + y^2)/(x - y)` is a homogeneous function of degree 1.
A homogeneous differential equation of the `(dx)/(dy) = h(x/y)` can be solved by making the substitution.
Find the general solution of the differential equation:
(xy – x2) dy = y2 dx
Read the following passage:
|
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
Based on the above, answer the following questions:
- Show that (x2 – y2) dx + 2xy dy = 0 is a differential equation of the type `dy/dx = g(y/x)`. (2)
- Solve the above equation to find its general solution. (2)
A function \[F(x,y)\] is homogeneous of degree \[n\] when which condition holds?
Which substitutions are chosen for a homogeneous differential equation?
What form should the equation be reduced to before integration?
What replacement obtains the final answer after using a homogeneous substitution?
For \[F(x,y)=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], what is \[F(\lambda x,\lambda y)\]?
After putting \[y=vx\] in \[\frac{dy}{dx}=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], which equation results?
From \[v+x\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}\], what is \[x\frac{dv}{dx}\]?
Which final answer results after replacing \[v=\frac{y}{x}\] in \[\sin v=\ln|Cx|\]?
Which sequence correctly states the essential actions for obtaining the final answer of a homogeneous differential equation?
