English

Solve the following differential equation. dr + (2r)dθ= 8dθ - Mathematics and Statistics

Advertisements
Advertisements

Question

Solve the following differential equation.

dr + (2r)dθ= 8dθ

Sum
Advertisements

Solution

dr + (2r)dθ= 8dθ

`(dr)/(dθ)` + 2r = 8

The given equation is of the form

`(dr)/(dθ) + Pr = Q`

where, P = 2 and Q = 8

I.F. = `e ^(int^(P^dθ) = e^(int^(2^dθ) = e^(2θ)`

Solution of the given equation is

`r(I.F.) = int Q (I.F.) dθ + c`

`re^(2θ) = int 8 e^(2θ)  dθ + c`

`re^(2θ) = 8 int  e^(2θ)  dθ + c`

`re ^(2θ) = 8e^(2θ)/2 + c`

`re ^(2θ) = 4e^(2θ) + 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 1.8 | Page 168

RELATED QUESTIONS

Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].


Show that y = AeBx is a solution of the differential equation

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

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

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

(ey + 1) cos x dx + ey sin x dy = 0


\[x\frac{dy}{dx} + \cot y = 0\]

Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]


\[\frac{dy}{dx} = y \sin 2x, y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 , y\left( 0 \right) = 1\]

If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).


\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

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

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

Solve the following initial value problem:-

\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]


The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.


If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?


Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

 

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.


Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.


The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.


Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is ex, then its solution is


Solve:

(x + y) dy = a2 dx


Solve the following differential equation

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 


Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is


Solution of `x("d"y)/("d"x) = y + x tan  y/x` is `sin(y/x)` = cx


Solve the differential equation

`x + y dy/dx` = x2 + y2


Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×