Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\cos x\frac{dy}{dx} - \cos 2x = \cos 3x\]
\[ \Rightarrow dy = \frac{\cos 3x + \cos 2x}{\cos x}dx\]
\[ \Rightarrow dy = \frac{4 \cos^3 x - 3\cos x + 2 \cos^2 x - 1}{\cos x}dx\]
\[ \Rightarrow dy = \left( 4 \cos^2 x - 3 + 2\cos x - \sec x \right)dx\]
\[ \Rightarrow dy = \left[ 2\left( 2 \cos^2 x - 1 \right) - 1 + 2\cos x - \sec x \right]dx\]
\[ \Rightarrow dy = \left( 2\cos 2x - 1 + 2\cos x - \sec x \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( 2\cos 2x - 1 + 2\cos x - \sec x \right)dx\]
\[ \Rightarrow y = \sin 2x - x + 2\sin x - \log\left| \sec x + \tan x \right| + C\]
\[\text{ Hence, } y = \sin 2x - x + 2\sin x - \log\left| \sec x + \tan x \right| +\text{C is the solution to the given differential equation.}\]
APPEARS IN
संबंधित प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
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 \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
(sin x + cos x) dy + (cos x − sin x) dx = 0
x cos2 y dx = y cos2 x dy
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
(y2 − 2xy) dx = (x2 − 2xy) dy
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
`dy/dx + 2xy = x`
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
Solve the differential equation:
dr = a r dθ − θ dr
Solve
`dy/dx + 2/ x y = x^2`
Select and write the correct alternative from the given option for the question
Differential equation of the function c + 4yx = 0 is
The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 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`
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
Solve the differential equation
`y (dy)/(dx) + x` = 0
The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.
Why is the equation \[x\frac{dy}{dx} + y = 0\] classified as a differential equation?
