Advertisements
Advertisements
Question
Advertisements
Solution
\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = \tan x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\tan x dx\]
\[ \Rightarrow \log \left| y \right| = \log \left| \sec x \right| + C . . . . . (1)\]
We know that at x = 0 and y = 1 .
Substituting the values of x and y in (1), we get
\[\log \left| 1 \right| = \log \left| 1 \right| + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[\log \left| y \right| = \log \left| \sec x \right| + 0\]
\[ \Rightarrow y = \sec x\]
\[\text{ Hence, }y = \sec x,\text{ where }x \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right),\text{ is the required solution .}\]
APPEARS IN
RELATED QUESTIONS
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 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 = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
y (1 + ex) dy = (y + 1) ex dx
(y2 + 1) dx − (x2 + 1) dy = 0
x2 dy + y (x + y) dx = 0
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
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 \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
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` |
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Select and write the correct alternative from the given option for the question
The differential equation of y = Ae5x + Be–5x is
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
Solve the differential equation
`x + y dy/dx` = x2 + y2
