English

Solve the following differential equation. (x+a)dydx=–y+a - Mathematics and Statistics

Advertisements
Advertisements

Question

Solve the following differential equation.

`(x + a) dy/dx = – y + a`

Sum
Advertisements

Solution

`(x + a) dy/dx = – y + a`

∴ `dy/dx + y/((x+a)) = a / ((x+a))`

The given equation is of the form

`dy/ dx + py = Q`

where, `P = 1/((x+a)) and Q = a/((x+a))`

∴ I.F. = `e ^(int^(pdx) = e ^(int^(1/(x+a))^dx)`

= `e^(log^ |x+a|) = (x+a)` 

∴ Solution of the given equation is

`y ( I.F.) = int Q (I.F.) dx + c `

∴ `y(x + a) = int a/((x+a)) (x+a) dx + c`

∴ `y(x + a) = a int 1  dx + c`

∴ y (x + a) = ax + c

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Differential Equation and Applications - Exercise 8.5 [Page 168]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.5 | Q 1.7 | Page 168

RELATED QUESTIONS

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.


Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]


Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]


Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]


Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex


\[\frac{dy}{dx} = x^5 + x^2 - \frac{2}{x}, x \neq 0\]

(sin x + cos x) dy + (cos x − sin x) dx = 0


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

\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]

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

(y2 + 1) dx − (x2 + 1) dy = 0


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

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

x2 dy + y (x + y) dx = 0


Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


Solve the following initial value problem:-

\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]


Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.


Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\]  are rectangular hyperbola.


Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]


The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is


y2 dx + (xy + x2)dy = 0


 `dy/dx = log x`


y dx – x dy + log x dx = 0


Solve the differential equation xdx + 2ydy = 0


Choose the correct alternative:

Differential equation of the function c + 4yx = 0 is


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


Solve the differential equation `"dy"/"dx" + 2xy` = y


A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×