Advertisements
Advertisements
प्रश्न
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`.
Advertisements
उत्तर
Given equation is `(1 + "t")"dy"/"dt" - "t"y` = 1
⇒ `"dy"/"dt" - ("t"/(1 + "t")) y = 1/(1 + "t")`
Here, P = `(-"t")/(1 + "t")` and Q = `1/(1 + "t")`
∴ Integrating factor I.F. = `"e"^(intpdt)`
= `"e"^(int (-1)/(1 + "t") "dt")`
= `"e"^(-int (1 + "t" - 1)/(1 + "t") "dt")`
= `"e"^(-int(1 - 1/(1 + "t"))"dt")`
= `"e"^(-["t" - log(1 + "t")])`
= `"e"^(-"t" + log(1 + "t"))`
= `"e"^(-"t") * "e"^(log(1 + "t"))`
∴ I.F. = `"e"^(-"t") * (1 + "t")`
Required solution of the given differential equation is
y . I. F. = `int "Q" . "I"."F". "dt" + "c"`
⇒ `y * "e"^-"t" (1 + "t") = int 1/((1 + "t")) * "e"^-"t" * (1 + "t") "dt" + "c"`
⇒ `y * "e"^-"t" (1 + "t") = int "e"^-"t" "dt" + "c"`
⇒ `y * "e"^-"t" (1 + "t") = - "e"^-"t" + "c"`
Put t = 0 and y = –1 ....[∵ y(0) = –1]
⇒ `-1 * "e"^0 * 1 = -"e"^0 + "c"`
⇒ –1 = –1 + c
⇒ c = 0
So the equation becomes
`y"e"^-"t" (1 + "t") = -"e"^-"t"`
Now put t = 1
∴ `y * "e"^-1 (1 + 1) = -"e"^-1`
⇒ 2y = –1
⇒ y = `- 1/2`
Hence y(1) = `-1/2` is verified.
APPEARS IN
संबंधित प्रश्न
Solve the differential equation: `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 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 the differential equation x (1 + y2) dx – y (1 + x2) dy = 0, given that y = 1 when x = 0.
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = x sin x : xy' = `y + x sqrt (x^2 - y^2)` (x ≠ 0 and x > y or x < -y)
Solve the differential equation `[e^(-2sqrtx)/sqrtx - y/sqrtx] dx/dy = 1 (x != 0).`
Find a particular solution of the differential equation`(x + 1) dy/dx = 2e^(-y) - 1`, given that y = 0 when x = 0.
If y = etan x+ (log x)tan x then find dy/dx
How many arbitrary constants are there in the general solution of the differential equation of order 3.
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
The solution of the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\], is
The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is
Which of the following differential equations has y = x as one of its particular solution?
The general solution of a differential equation of the type \[\frac{dx}{dy} + P_1 x = Q_1\] 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
The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is
(x + y − 1) dy = (x + y) dx
(x2 + 1) dy + (2y − 1) dx = 0
\[\frac{dy}{dx} + y = 4x\]
\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]
\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]
Solve the differential equation:
(1 + y2) dx = (tan−1 y − x) dy
Solve the following differential equation:- \[x \cos\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]
Solve the following differential equation:- `y dx + x log (y)/(x)dy-2x dy=0`
Solve the following differential equation:-
\[\frac{dy}{dx} + 2y = \sin x\]
Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is `(2x)/y^2.`
The general solution of the differential equation `"dy"/"dx" + y sec x` = tan x is y(secx – tanx) = secx – tanx + x + k.
Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
Solve:
`2(y + 3) - xy (dy)/(dx)` = 0, given that y(1) = – 2.
Find the general solution of (1 + tany)(dx – dy) + 2xdy = 0.
The general solution of ex cosy dx – ex siny dy = 0 is ______.
y = aemx+ be–mx satisfies which of the following differential equation?
The general solution of the differential equation `("d"y)/("d"x) = "e"^(x^2/2) + xy` is ______.
The differential equation for which y = acosx + bsinx is a solution, is ______.
The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is ______.
The integrating factor of `("d"y)/("d"x) + y = (1 + y)/x` is ______.
The member of arbitrary constants in the particulars solution of a differential equation of third order as
Find the general solution of the differential equation `x (dy)/(dx) = y(logy - logx + 1)`.
