Advertisements
Advertisements
Question
Solve the following differential equation.
dr + (2r)dθ= 8dθ
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`
APPEARS IN
RELATED QUESTIONS
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
y (1 + ex) dy = (y + 1) ex dx
dy + (x + 1) (y + 1) dx = 0
Solve the following initial value problem:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
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 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 solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by
The solution of the differential equation y1 y3 = y22 is
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
In each of the following examples, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = ex | `dy/ dx= y` |
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| ax2 + by2 = 5 | `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx` |
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
