Advertisements
Advertisements
प्रश्न
Solve the following differential equation:- `y dx + x log (y)/(x)dy-2x dy=0`
Advertisements
उत्तर
We have,
\[y dx + x \log \left( \frac{y}{x} \right)dy - 2x dy = 0\]
\[ \Rightarrow x \log \left( \frac{y}{x} \right)dy - 2x dy = - y dx\]
\[ \Rightarrow \left[ \log \left( \frac{y}{x} \right) - 2 \right]x dy = - y dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{\left[ \log \left( \frac{y}{x} \right) - 2 \right]x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)} . . . . . . . . \left( 1 \right)\]
Clearly this is a homogenous equation,
Putting y = vx
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{Substituting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ in (1) we get}\]
\[v + x\frac{dv}{dx} = \frac{v}{2 - \log \left( v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v}{2 - \log \left( v \right)} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v - 2v + v \log \left( v \right)}{2 - \log \left( v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- v + v \log \left( v \right)}{2 - \log \left( v \right)}\]
\[ \Rightarrow \frac{2 - \log \left( v \right)}{- v + v \log \left( v \right)}dv = \frac{1}{x}dx\]
\[ \Rightarrow \frac{\log \left( v \right) - 2}{v \log \left( v \right) - v}dv = - \frac{1}{x}dx\]
\[ \Rightarrow \frac{\log \left( v \right) - 1 - 1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]
\[ \Rightarrow \frac{\log \left( v \right) - 1}{v \left[ \log \left( v \right) - 1 \right]}dv - \frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{v}dv - \frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]
Integrating both sides we get
\[\int\frac{1}{v}dv - \int\frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| v \right| - I = - \log \left| x \right| - \log C . . . . . . . . \left( 2 \right)\]
Where,
\[I = \int\frac{1}{v \left[ \log \left| \left( v \right) \right| - 1 \right]}dv\]
Puting log v = t
\[\frac{1}{v}dv = dt\]
\[ \therefore I = \int\frac{1}{t - 1}dt\]
\[ \Rightarrow I = \log \left| t - 1 \right|\]
\[ \Rightarrow I = \log \left| \log \left| v \right| - 1 \right| . . . . . \left( 3 \right)\]
From (2) and (3) we get
\[\log \left| v \right| - \log \left| \log \left| v \right| - 1 \right| = - \log \left| x \right| - \log C\]
\[ \Rightarrow \log \left| \frac{v}{\log \left| v \right| - 1} \right| = - \log \left| Cx \right|\]
\[ \Rightarrow \frac{v}{\log \left| v \right| - 1} = \frac{1}{Cx}\]
\[ \Rightarrow \log \left| v \right| - 1 = vCx\]
\[ \Rightarrow \log \left| \frac{y}{x} \right| - 1 = Cy\]
APPEARS IN
संबंधित प्रश्न
The differential equation of `y=c/x+c^2` is :
(a)`x^4(dy/dx)^2-xdy/dx=y`
(b)`(d^2y)/dx^2+xdy/dx+y=0`
(c)`x^3(dy/dx)^2+xdy/dx=y`
(d)`(d^2y)/dx^2+dy/dx-y=0`
The solution of the differential equation dy/dx = sec x – y tan x is:
(A) y sec x = tan x + c
(B) y sec x + tan x = c
(C) sec x = y tan x + c
(D) sec x + y tan x = c
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
Find the particular solution of the differential equation `(1+x^2)dy/dx=(e^(mtan^-1 x)-y)` , give that y=1 when x=0.
Find the particular solution of the differential equation `dy/dx=(xy)/(x^2+y^2)` given that y = 1, when x = 0.
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = Ax : xy′ = y (x ≠ 0)
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
xy = log y + C : `y' = (y^2)/(1 - xy) (xy != 1)`
Show that the general solution of the differential equation `dy/dx + (y^2 + y +1)/(x^2 + x + 1) = 0` is given by (x + y + 1) = A (1 - x - y - 2xy), where A is parameter.
Solve the differential equation `[e^(-2sqrtx)/sqrtx - y/sqrtx] dx/dy = 1 (x != 0).`
The population of a town grows at the rate of 10% per year. Using differential equation, find how long will it take for the population to grow 4 times.
Solve the differential equation:
`e^(x/y)(1-x/y) + (1 + e^(x/y)) dx/dy = 0` when x = 0, y = 1
Write the order of the differential equation associated with the primitive y = C1 + C2 ex + C3 e−2x + C4, where C1, C2, C3, C4 are arbitrary constants.
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is
\[\frac{dy}{dx} + \frac{y}{x} = \frac{y^2}{x^2}\]
\[\frac{dy}{dx} - y \tan x = e^x \sec x\]
(1 + y + x2 y) dx + (x + x3) dy = 0
x2 dy + (x2 − xy + y2) dx = 0
Solve the differential equation:
(1 + y2) dx = (tan−1 y − x) dy
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sin^{- 1} x\]
Solve the following differential equation:-
\[\frac{dy}{dx} + 3y = e^{- 2x}\]
Solve the following differential equation:-
\[\left( x + y \right)\frac{dy}{dx} = 1\]
Solve the following differential equation:-
\[\left( x + 3 y^2 \right)\frac{dy}{dx} = y\]
Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]
Find a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1
If y(x) is a solution of `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx and y (0) = 1, then find the value of `y(pi/2)`.
Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
Solution of differential equation xdy – ydx = 0 represents : ______.
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
The solution of the differential equation `("d"y)/("d"x) + (1 + y^2)/(1 + x^2)` is ______.
The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is ______.
General solution of the differential equation of the type `("d"x)/("d"x) + "P"_1x = "Q"_1` is given by ______.
Polio drops are delivered to 50 K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of 2nd week half the children have been given the polio drops. How many will have been given the drops by the end of 3rd week can be estimated using the solution to the differential equation `"dy"/"dx" = "k"(50 - "y")` where x denotes the number of weeks and y the number of children who have been given the drops.
The value of c in the particular solution given that y(0) = 0 and k = 0.049 is ______.
Find the general solution of the differential equation `x (dy)/(dx) = y(logy - logx + 1)`.
Find the general solution of the differential equation:
`log((dy)/(dx)) = ax + by`.
If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then ______.
