Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\left( x - y \right) \frac{dy}{dx} = x + 2y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y}\]
This is a homogeneous differential equatiuon .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}, \text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v - v + v^2}{1 - v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + v + v^2}{1 - v}\]
\[ \Rightarrow \frac{1 - v}{1 + v + v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1 - v}{1 + v + v^2}dv = \int\frac{1}{x}dx\]
\[\Rightarrow \int\frac{- (v - 1)}{v^2 + v + 1}dv = \frac{dx}{x}\]
\[ \Rightarrow \int\frac{1}{2} \times \frac{2v - 2}{v^2 + v + 1}dv = \int\frac{- dx}{x}\]
\[ \Rightarrow \int\frac{(2v + 1) - 3}{v^2 + v + 1}dv = - \int\frac{2dx}{x}\]
\[ \Rightarrow \int\frac{(2v + 1)}{v^2 + v + 1}dv - \int\frac{3}{v^2 + v + 1}dv = - \int\frac{2dx}{x}\]
\[\text{ Let }I_1 = \int\frac{(2v + 1)}{v^2 + v + 1}dv\]
\[\text{ and }I_2 = \int\frac{3}{v^2 + v + 1}dv\]
\[I = I_1 + I_2 \]
\[I_1 = \int \frac{2v + 1}{v^2 + v + 1}dv\]
\[\text{ let }v^2 + v + 1 = t \Rightarrow (2 v^2 + 1)dv = dt\]
\[\text{ therefore, }I_1 = \int \frac{2v + 1}{v^2 + v + 1}dv = \int\frac{dt}{t} = \log\left| t \right| = \log\left| v^2 + v + 1 \right|\]
\[\text{ hence, } I_1 = log\left| v^2 + v + 1 \right|\]
\[\text{ Also }, I_2 = \int\frac{3}{v^2 + v + 1}dv = \int\frac{3}{v^2 + 2v\left( \frac{1}{2} \right) + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1}\]
\[ = \int\frac{3}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dv = 3\left( \frac{2}{\sqrt{3}} \right) \tan^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[ I_2 = 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[\text{ Hence, }I = I_1 + I_2 = log\left| v^2 + v + 1 \right| + 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[\text{ therefore, }log\left| v^2 + v + 1 \right| + 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = - 2\log\left| x \right| + C\]
putting the value of v in the above equation we get,
\[log\left| x^2 + y^2 + xy \right| = 2\sqrt{3} ta n^{- 1} \left( \frac{x + 2y}{x\sqrt{3}} \right) + C\]
APPEARS IN
संबंधित प्रश्न
Solve the differential equation (x2 + y2)dx- 2xydy = 0
Solve the differential equation :
`y+x dy/dx=x−y dy/dx`
Find the particular solution of the differential equation:
2y ex/y dx + (y - 2x ex/y) dy = 0 given that x = 0 when y = 1.
Show that the given differential equation is homogeneous and solve them.
(x2 + xy) dy = (x2 + y2) dx
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.
(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.
`{xcos(y/x) + ysin(y/x)}ydx = {ysin (y/x) - xcos(y/x)}xdy`
Show that the given differential equation is homogeneous and solve them.
`x dy/dx - y + x sin (y/x) = 0`
Show that the given differential equation is homogeneous and solve them.
`y dx + x log(y/x)dy - 2x dy = 0`
For the differential equation find a particular solution satisfying the given condition:
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
For the differential equation find a particular solution satisfying the given condition:
`2xy + y^2 - 2x^2 dy/dx = 0; y = 2` when x = 1
A homogeneous differential equation of the from `dx/dy = h (x/y)` can be solved by making the substitution.
Which of the following is a homogeneous differential equation?
(2x2 y + y3) dx + (xy2 − 3x3) dy = 0
Solve the following initial value problem:
\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y\left( 2 \right) = x\]
Find the particular solution of the differential equation x cos\[\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\], given that when x = 1, \[y = \frac{\pi}{4}\]
Find the particular solution of the differential equation \[\left( x - y \right)\frac{dy}{dx} = x + 2y\], given that when x = 1, y = 0.
Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]
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:
`x * dy/dx - y + x * sin(y/x) = 0`
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
Solve the following differential equation:
`x^2. dy/dx = x^2 + xy + y^2`
Find the equation of a curve passing through `(1, pi/4)` if the slope of the tangent to the curve at any point P(x, y) is `y/x - cos^2 y/x`.
F(x, y) = `(sqrt(x^2 + y^2) + y)/x` is a homogeneous function of degree ______.
F(x, y) = `(ycos(y/x) + x)/(xcos(y/x))` is not a homogeneous function.
The differential equation y' = `y/(x + sqrt(xy))` has general solution given by:
(where C is a constant of integration)
