English

Solve the following differential equation. y2 dx + (xy + x2 ) dy = 0

Advertisements
Advertisements

Question

Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0

Sum
Advertisements

Solution

y2 dx + (xy + x2 ) dy = 0

∴ (xy + x2 ) dy = - y2 dx

∴`dy/dx = (-y^2)/(xy+x^2)`  ...(i)

Put y = tx  ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t + x dt/dx`  ...(iii)

Substituting (ii) and (iii) in (i), we get

`t + x dt/dx  = (-t^2x^2)/(x.tx+x^2)` 

∴`t + x dt/dx  = (-t^2x^2)/(tx^2+x^2)`

∴  `t + x dt/dx  = (-t^2x^2)/(x^2(t+1)`

∴  ` x dt/dx  = (-t^2)/(t+1)-t`

∴ ` x dt/dx  = (-t^2-t^2-t)/(t+1)`

∴ ` x dt/dx  = (-(2t^2+t))/(t+1)`

∴ `(t+1)/(2t^2+t)dt = - 1/x dx`

Integrating on both sides, we get

`int (t+1)/(2t^2+t)dt = -int1/xdx`

∴`int (2t + 1 - t)/(t(2t+1)) dt = - int1/xdx`

∴`int1/tdt-int  1/ (2t+1) dt = -int1/ x dx`

∴ log | t | - `1/2` log |2t + 1| = - log |x| + log |c|

∴ 2log| t | - log |2t + 1| = - 2log |x| + 2 log |c|

∴ `2log |y/x | -log |(2y)/x+ 1 |= - 2log |x| + 2 log |c|`

∴ 2log |y| - 2log |x| - log |2y + x| + log |x|

= -2log |x| + 2log |c|

∴ log |y2 | + log |x| = log |c2 | + log |2y + x|

∴ log |y2 x| = log | c2 (x + 2y)|

∴ xy 2 = c2 (x + 2y)

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Differential Equation and Applications - Exercise 8.4 [Page 167]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.4 | Q 1.2 | Page 167

RELATED QUESTIONS

\[x + \left( \frac{dy}{dx} \right) = \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]

Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]


Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x\frac{dy}{dx} = y\]
y = ax

\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

\[\frac{dy}{dx} = \sin^2 y\]

\[\left( x - 1 \right)\frac{dy}{dx} = 2 xy\]

(1 + x2) dy = xy dx


tan y dx + sec2 y tan x dy = 0


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


(y + xy) dx + (x − xy2) dy = 0


\[\frac{dy}{dx} = 1 - x + y - xy\]

\[xy\frac{dy}{dx} = y + 2, y\left( 2 \right) = 0\]

\[\frac{dy}{dx} = \left( x + y + 1 \right)^2\]

\[\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\]

\[\frac{dy}{dx} = \frac{x + y}{x - y}\]

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.


Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\]  at any point (x, y) on it.


Define a differential equation.


y2 dx + (x2 − xy + y2) dy = 0


Form the differential equation representing the family of curves y = a sin (x + b), where ab are arbitrary constant.


Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.


Determine the order and degree of the following differential equations.

Solution D.E
y = aex + be−x `(d^2y)/dx^2= 1`

Solve the following differential equation.

`(x + a) dy/dx = – y + a`


The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.


Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×