Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[x\frac{dy}{dx} = y - x \cos^2 \left( \frac{y}{x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y - x \cos^2 \left( \frac{y}{x} \right)}{x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = v - \cos^2 v\]
\[ \Rightarrow x\frac{dv}{dx} = - \cos^2 v\]
\[ \Rightarrow \frac{1}{\cos^2 v}dv = - \frac{1}{x}dx\]
\[ \Rightarrow \sec^2 v = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int \sec^2 v dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \tan v = - \log \left| x \right| + \log C \]
\[ \Rightarrow \tan v = \log \left| \frac{C}{x} \right|\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[\tan \left( \frac{y}{x} \right) = \log \left| \frac{C}{x} \right|\]
\[ \Rightarrow \tan \left( \frac{y}{x} \right) = \log \left| \frac{C}{x} \right|\text{ is the required solution }.\]
APPEARS IN
संबंधित प्रश्न
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 1\] Function y = sin x + cos x
xy dy = (y − 1) (x + 1) dx
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
(y + xy) dx + (x − xy2) dy = 0
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
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).
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.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Define a differential equation.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by
The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
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` |
For each of the following differential equations find the particular solution.
`y (1 + logx)dx/dy - x log x = 0`,
when x=e, y = e2.
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
Solve the following differential equation.
`dy/dx + y = e ^-x`
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
Solve the differential equation
`x + y dy/dx` = x2 + y2
