हिंदी

If y(t) is a solution of tdydtt(1+t)dydt-ty = 1 and y(0) = – 1, then show that y(1) = -12.

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`.

योग
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.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Exercise [पृष्ठ १९३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 9 Differential Equations
Exercise | Q 12 | पृष्ठ १९३

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

The solution of the differential equation dy/dx = sec x – y tan x is:

(A) y sec x = tan x + c

(B) y sec x + tan x = c

(C) sec x = y tan x + c

(D) sec x + y tan x = c


Solve the differential equation:  `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 0.


If x = Φ(t) differentiable function of ‘ t ' then prove that `int f(x) dx=intf[phi(t)]phi'(t)dt`


Solve : 3ex tanydx + (1 +ex) sec2 ydy = 0

Also, find the particular solution when x = 0 and y = π.


If y = P eax + Q ebx, show that

`(d^y)/(dx^2)=(a+b)dy/dx+aby=0`


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan–1y   :   y2 y′ + y2 + 1 = 0


Find a particular solution of the differential equation`(x + 1) dy/dx = 2e^(-y) - 1`, given that y = 0 when x = 0.


Find `(dy)/(dx)` at x = 1, y = `pi/4` if `sin^2 y + cos xy = K`


How many arbitrary constants are there in the general solution of the differential equation of order 3.


The general solution of the differential equation \[\frac{dy}{dx} + y \] cot x = cosec x, is


If m and n are the order and degree of the differential equation \[\left( y_2 \right)^5 + \frac{4 \left( y_2 \right)^3}{y_3} + y_3 = x^2 - 1\], then


The general solution of a differential equation of the type \[\frac{dx}{dy} + P_1 x = Q_1\] is


The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is


Solve the differential equation (x2 − yx2) dy + (y2 + x2y2) dx = 0, given that y = 1, when x = 1.


\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]


\[\frac{dy}{dx} - y \tan x = e^x \sec x\]


`(2ax+x^2)(dy)/(dx)=a^2+2ax`


\[y - x\frac{dy}{dx} = b\left( 1 + x^2 \frac{dy}{dx} \right)\]


\[\frac{dy}{dx} + 5y = \cos 4x\]


Find the general solution of the differential equation \[\frac{dy}{dx} = \frac{x + 1}{2 - y}, y \neq 2\]


For the following differential equation, find a particular solution satisfying the given condition:

\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]


Solve the following differential equation:- `y dx + x log  (y)/(x)dy-2x dy=0`


Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]


Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.


The solution of the differential equation `x "dt"/"dx" + 2y` = x2 is ______.


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)`.


Solve:

`2(y + 3) - xy  (dy)/(dx)` = 0, given that y(1) = – 2.


Solve the differential equation dy = cosx(2 – y cosecx) dx given that y = 2 when x = `pi/2`


The number of solutions of `("d"y)/("d"x) = (y + 1)/(x - 1)` when y (1) = 2 is ______. 


The general solution of ex cosy dx – ex siny dy = 0 is ______.


y = aemx+ be–mx satisfies which of the following differential equation?


The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is ______.


The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is ______.


The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.


General solution of the differential equation of the type `("d"x)/("d"x) + "P"_1x = "Q"_1` is given by ______.


General solution of `("d"y)/("d"x) + y` = sinx is ______.


Find the general solution of the differential equation `x (dy)/(dx) = y(logy - logx + 1)`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×