Advertisements
Advertisements
प्रश्न
(sin x + cos x) dy + (cos x − sin x) dx = 0
Advertisements
उत्तर
We have,
\[\left( \sin x + \cos x \right)dy + \left( \cos x - \sin x \right)dx = 0\]
\[ \Rightarrow dy = - \left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
Integrating both sides, we get
\[\int dy = - \int\left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
\[ \Rightarrow y = - \int\left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
\[\text{ Putting }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]
\[ \therefore y = - \int\frac{dt}{t}\]
\[ \Rightarrow y = - \log\left| t \right| + C\]
\[ \Rightarrow y = - \log\left| \sin x + \cos x \right| + C\]
\[ \Rightarrow y + \log\left| \sin x + \cos x \right| = C\]
\[\text{ Hence, }y + \log\left| \sin x + \cos x \right| =\text{ C is the solution to the given differential equation }.\]
APPEARS IN
संबंधित प्रश्न
If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega + b omega^2) = omega^2`
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Show that y = AeBx is a solution of the differential equation
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
(y + xy) dx + (x − xy2) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
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).
3x2 dy = (3xy + y2) dx
Solve the following initial value problem:-
\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]
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:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
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.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
The differential equation satisfied by ax2 + by2 = 1 is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
y2 dx + (x2 − xy + y2) dy = 0
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
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.
(x2 − y2 ) dx + 2xy dy = 0
Solve the differential equation:
`e^(dy/dx) = x`
Solve the differential equation:
dr = a r dθ − θ dr
`xy dy/dx = x^2 + 2y^2`
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
