Advertisements
Advertisements
प्रश्न
Solve the following initial value problem:
(xy − y2) dx − x2 dy = 0, y(1) = 1
Advertisements
उत्तर
(xy − y2) dx − x2 dy = 0, y(1) = 1
This is an homogenous equation, put y = vx
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\left( xy - y^2 \right) = x^2 \left( \frac{dy}{dx} \right)\]
\[\left( v x^2 - v^2 x^2 \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v x^2 \left( 1 - v \right) = x^2 \left( v + x\frac{dv}{dx} \right)\]
\[v\left( 1 - v \right) = v + x\frac{dv}{dx}\]
\[v - v^2 = v + x\frac{dv}{dx}\]
\[ - v^2 = x\frac{dv}{dx}\]
\[ - \frac{1}{x}dx = \frac{1}{v^2}dv\]
On integrating both sides we get,
\[- \int\frac{1}{x}dx = \int\frac{1}{v^2}dv\]
\[ - \log_e x = \frac{v^{- 2 + 1}}{- 2 + 1} + c\]
\[ - \log_e x = \frac{v^{- 1}}{- 1} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[ - \log_e x = - \frac{1}{v} + c\]
\[\frac{x}{y} - \log_e x = c\]
\[\text{ As }y\left( 1 \right) = 1\]
\[\frac{1}{1} - \log_e 1 = c\]
\[ \Rightarrow c = 1\]
APPEARS IN
संबंधित प्रश्न
Solve the differential equation (x2 + y2)dx- 2xydy = 0
Show that the given differential equation is homogeneous and solve them.
(x2 – y2) dx + 2xy dy = 0
Show that the given differential equation is homogeneous and solve them.
`x^2 dy/dx = x^2 - 2y^2 + xy`
Show that the given differential equation is homogeneous and solve them.
`y dx + x log(y/x)dy - 2x dy = 0`
Show that the given differential equation is homogeneous and solve them.
`(1+e^(x/y))dx + e^(x/y) (1 - x/y)dy = 0`
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`
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
(x2 + 3xy + y2) dx − x2 dy = 0
Solve the following initial value problem:
x (x2 + 3y2) dx + y (y2 + 3x2) dy = 0, y (1) = 1
Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]
A homogeneous differential equation of the form \[\frac{dx}{dy} = h\left( \frac{x}{y} \right)\] can be solved by making the substitution
Solve the following differential equation : \[\left[ y - x \cos\left( \frac{y}{x} \right) \right]dy + \left[ y \cos\left( \frac{y}{x} \right) - 2x \sin\left( \frac{y}{x} \right) \right]dx = 0\] .
Solve the following differential equation:
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
Solve the following differential equation:
`(1 + "e"^("x"/"y"))"dx" + "e"^("x"/"y")(1 - "x"/"y")"dy" = 0`
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
Solve the following differential equation:
x dx + 2y dx = 0, when x = 2, y = 1
Solve the following differential equation:
`x^2. dy/dx = x^2 + xy + y^2`
State whether the following statement is True or False:
A homogeneous differential equation is solved by substituting y = vx and integrating it
Which of the following is not a homogeneous function of x and y.
F(x, y) = `(ycos(y/x) + x)/(xcos(y/x))` is not a homogeneous function.
A homogeneous differential equation of the `(dx)/(dy) = h(x/y)` can be solved by making the substitution.
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation, 2x2dy = (2xy + y2)dx, then `f(1/2)` is equal to ______.
The differential equation y' = `y/(x + sqrt(xy))` has general solution given by:
(where C is a constant of integration)
Find the general solution of the differential equation:
(xy – x2) dy = y2 dx
The solution of the equation `dy/dx = (3x − 4y − 2)/(3x − 4y − 3)` is ______.
Which differential equation is called a homogeneous differential equation?
After using \[y=vx\] and writing the right-hand side as \[g(v)\], which separable form is obtained?
What form should the equation be reduced to before integration?
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 \[x\frac{dv}{dx}\]?
