Advertisements
Advertisements
Question
\[\frac{dy}{dx} - y \tan x = e^x \sec x\]
Advertisements
Solution
We have,
\[\frac{dy}{dx} - y \tan x = e^x \sec x\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = - \tan x \]
\[Q = e^x \sec x\]
Now,
\[I . F . = e^{\int - \tan x\ dx} \]
\[ = e^{- \log\left| \left( \sec x \right) \right|} \]
\[ = e^{\log\left| \left( \cos x \right) \right|} \]
\[ = \cos x\]
So, the solution is given by
\[y \cos\ x = \int\left( \cos x\ e^x \sec x \right) dx + C\]
\[ \Rightarrow y \cos\ x = \int e^x dx + C\]
\[ \therefore y \cos\ x = e^x + C\]
APPEARS IN
RELATED QUESTIONS
The differential equation of `y=c/x+c^2` is :
(a)`x^4(dy/dx)^2-xdy/dx=y`
(b)`(d^2y)/dx^2+xdy/dx+y=0`
(c)`x^3(dy/dx)^2+xdy/dx=y`
(d)`(d^2y)/dx^2+dy/dx-y=0`
Find the particular solution of the differential equation `e^xsqrt(1-y^2)dx+y/xdy=0` , given that y=1 when x=0
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
`y sqrt(1 + x^2) : y' = (xy)/(1+x^2)`
The number of arbitrary constants in the particular solution of a differential equation of third order are ______.
The general solution of the differential equation \[\frac{dy}{dx} + y\] g' (x) = g (x) g' (x), where g (x) is a given function of x, is
The solution of the differential equation \[\frac{dy}{dx} - ky = 0, y\left( 0 \right) = 1\] approaches to zero when x → ∞, if
The general solution of the differential equation \[\frac{dy}{dx} = e^{x + y}\], is
The general solution of the differential equation \[\frac{y dx - x dy}{y} = 0\], is
Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{x\left( 2 \log x + 1 \right)}{\sin y + y \cos y}\] given that
x (e2y − 1) dy + (x2 − 1) ey dx = 0
\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]
\[\frac{dy}{dx} - y \tan x = e^x\]
(1 + y + x2 y) dx + (x + x3) dy = 0
`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`
\[\frac{dy}{dx} + 2y = \sin 3x\]
\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]
`x cos x(dy)/(dx)+y(x sin x + cos x)=1`
For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]
Solve the following differential equation:- \[\left( x - y \right)\frac{dy}{dx} = x + 2y\]
Solve the following differential equation:-
\[x\frac{dy}{dx} + 2y = x^2 , x \neq 0\]
Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]
Find a particular solution of the following differential equation:- x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
Solve the differential equation: `(d"y")/(d"x") - (2"x")/(1+"x"^2) "y" = "x"^2 + 2`
Solve the differential equation : `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0 "given that" "y" = 0 "when" "x" = 1`.
Solution of the differential equation `"dx"/x + "dy"/y` = 0 is ______.
The general solution of the differential equation x(1 + y2)dx + y(1 + x2)dy = 0 is (1 + x2)(1 + y2) = k.
Find the general solution of `(x + 2y^3) "dy"/"dx"` = y
If y(x) is a solution of `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx and y (0) = 1, then find the value of `y(pi/2)`.
If y(t) is a solution of `(1 + "t")"dy"/"dt" - "t"y` = 1 and y(0) = – 1, then show that y(1) = `-1/2`.
Find the general solution of the differential equation `(1 + y^2) + (x - "e"^(tan - 1y)) "dy"/"dx"` = 0.
Solve the differential equation (1 + y2) tan–1xdx + 2y(1 + x2)dy = 0.
Solution of differential equation xdy – ydx = 0 represents : ______.
General solution of `("d"y)/("d"x) + ytanx = secx` is ______.
The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.
The member of arbitrary constants in the particulars solution of a differential equation of third order as
