Find the general solution of each of the following differential equations:
Rearranging the terms,we get:
Integrating both the sides we get,
…()
Ans:
Find the general solution of each of the following differential equations:
Integrating both the sides we get,
…(3c’ = c)
Find the general solution of each of the following differential equations:
Rearranging the terms we get:
Integrating both the sides we get,
…(
Ans:
Find the general solution of each of the following differential equations:
Rearranging the terms we get:
Integrating both the sides we get,
…(
Ans:
Find the general solution of each of the following differential equations:
Separating the variables we get:
Integrating both the sides we get,
Ans:
Find the general solution of each of the following differential equations:
Rearringing the terms we get:
Integrating both the sides we get,
Ans:ex + e - y = c
Find the general solution of each of the following differential equations:
(ex + e - x )dy - (ex - e - x)dx = 0
Integrating both the sides we get,
y = log|ex + e - x| + c …(
Ans:y = log|ex + e - x| + c
Find the general solution of each of the following differential equations:
Given:
Integrating both the sides we get:
Ans:
Find the general solution of each of the following differential equations:
e2xe - 3ydx + e2ye - 3xdy = 0
Rearringing the terms we get:
e2x + 3xdx = - e2y + 3ydy
⇒e5xdx = - e5ydy
Integrating both the sides we get:
⇒e5x + e5y = 5c’ = c
Ans: e5x + e5y = c
Find the general solution of each of the following differential equations:
ex tan y dx + (1 – ex) sec2 y dy = 0
Rearranging all the terms we get:
Integrating both the sides we get:
⇒log|1 - ex| = log|tany| - logc
⇒log|1 - ex| + logc = log|tany|
⇒tany = c(1 - ex)
Ans: tany = c(1 - ex)
Find the general solution of each of the following differential equations:
sec2x tan y dx + sec2y tan x dy = 0
Rearranging the terms we get:
Integrating both the sides we get:
⇒log|tanx| = - log|tany| + logc
⇒ log|tanx| + log|tany| = logc
⇒tanx.tany = c
Ans: tanx.tany = c
Find the general solution of each of the following differential equations:
cos x(1 + cos y)dx – sin y(1 + sin x)dy = 0
Rearranging the terms we get:
Integrating both the sides we get:
⇒log|1 + sinx| = - log|1 + cosy| + logc
⇒log|1 + sinx| + log|1 + cosy| = logc
⇒(1 + sinx)(1 + cosy) = c
Ans: (1 + sinx)(1 + cosy) = c
For each of the following differential equations, find a particular solution satisfying the given condition :
.. where a ∈ R and y = 2 when x = 0.
⇒
⇒dy = cos - 1a dx
Integrating both the sides we get:
⇒y = xcos - 1a + c
when x = 0, y = 2
∴2 = 0 + c
∴c = 2
∴y = xcos - 1a + 2
Ans:
For each of the following differential equations, find a particular solution satisfying the given condition :
it being given that y = 1 when x = 0.
Rearranging the terms we get:
Integrating both the sides we get:
⇒y - 1 = 2x2 + c
y = 1 when x = 0
⇒ (1) - 1 = 2(0)2 + c
⇒c = 1
Ans:
For each of the following differential equations, find a particular solution satisfying the given condition :
x dy = (2x2 + 1) dx (x ≠ 0), given that
y = 1 when x = 1.
Rearranging the terms we get:
Integrating both the sides we get:
⇒y = x2 + log|x| + c
y = 1 when x = 1
∴1 = 12 + log1 + c
∴1 - 1 = 0 + c …(log1 = 0)
⇒c = 0
∴y = x2 + log|x|
Ans: y = x2 + log|x|
For each of the following differential equations, find a particular solution satisfying the given condition :
it being given that y = 1 when x = 0.
Rearranging the terms we get:
⇒log|y| = log|secx| + logc
⇒log|y| - log|secx| = logc
⇒log|y| + log|cosx| = logc
⇒ycosx = c
y = 1 when x = 0
∴1×cos0 = c
∴c = 1
⇒ycosx = 1
⇒y = 1/cosx
⇒y = secx
Ans: y = secx
Find the general solution of each of the following differential equations:
(y + 2)dy = (x - 1)dx
Integrating on both sides,
Find the general solution of each of the following differential equations:
Multiply and divide 2 in numerator and denominator of RHS,
Integrating on both sides
Find the general solution of each of the following differential equations:
Integrating on both sides
Find the general solution of each of the following differential equations:
Multiply and divide 2 in numerator and denominator of RHS,
Integrating on both sides
⇒1
Find the general solution of each of the following differential equations:
Integrating on both sides
Find the general solution of each of the following differential equations:
Integrating on both sides
Find the general solution of each of the following differential equations:
⟹
Integrating on both the sides,
LHS:
Let
Comparing coefficients in both the sides,
A = - 1, B = 1
RHS:
Therefore the solution of the given differential equation is
Find the general solution of each of the following differential equations:
x2 (y + 1) dx + y2 (x – 1) dy = 0
Add and subtract 1 in numerators of both LHS and RHS,
By the identity,
Splitting the terms,
Integrating,
Find the general solution of each of the following differential equations:
Multiply 2 in both LHS and RHS,
Integrating on both the sides,
Find the general solution of each of the following differential equations:
y log y dx – x dy = 0
Integrating on both the sides,
LHS:
RHS:
Let
So,
Therefore the solution of the given differential equation is
Find the general solution of each of the following differential equations:
x(x2 – x2 y2) dy + y(y2 + x2y2) dx = 0
Integrating ,
Find the general solution of each of the following differential equations:
(1 – x2) dy + xy (1 – y) dx = 0
Integrating on both the sides,
LHS:
Let
⇒
Comparing coefficients in both the sides,
A = - 1, B = 1
RHS:
Multiply and divide 2
Therefore the solution of the given differential equation is
-
=
Find the general solution of each of the following differential equations:
(1 – x2)(1 – y) dx = xy (1 + y) dy
Integrating on both the sides,
LHS:
RHS:
Add and subtract 1 in numerators of both LHS and RHS,
By the identity,
Splitting the terms,
Integrating,
Therefore the solution of the given differential equation is
Find the general solution of each of the following differential equations:
(y + xy) dx + (x – xy2) dy = 0
Integrating ,
Find the general solution of each of the following differential equations:
(x2 – yx2) dy + (y2 + xy2) dx = 0
Integrating,
Find the general solution of each of the following differential equations:
Add and subtract 1 in numerators ,
By the identity,
Splitting the terms,
Integrating,
Find the general solution of each of the following differential equations:
Integrating,
Find the general solution of each of the following differential equations:
Integrating on both the sides,
Find the general solution of each of the following differential equations:
Considering ‘d’ as exponential ’e’
Integrating on both the sides,
Find the general solution of each of the following differential equations:
⇒
Integrating on both the sides,
formula:
Find the general solution of each of the following differential equations:
Integrating on both the sides,
LHS:
By ILATE rule,
RHS:
Multiply and divide by 2
Therefore the solution of the given differential equation is
Find the general solution of each of the following differential equations:
Integrating on both the sides,
formula:
=
Find the general solution of each of the following differential equations:
(ey + 1) cos x dx + ey sin x dy = 0
Integrating,
Find the general solution of each of the following differential equations:
Integrating ,
Find the general solution of each of the following differential equations:
Multiply and divide by 2,
Integrating on both the sides,
formula:
⇒
Find the general solution of each of the following differential equations:
Integrating ,
Consider the integral
Let
So,
Consider the integral
By ILATE rule,
Again by ILATE rule,
Therefore the solution of the given differential equation is,
Find the general solution of each of the following differential equations:
Integrating,
Consider the integral
Let
So,
Consider the integral
Therefore the solution of the differential equation is
Find the general solution of each of the following differential equations:
Integrating on both the sides,
(adding and subtracting 1)
Find the general solution of each of the following differential equations:
Integrating,
Consider the integral
By ILATE rule,
Consider the integral
Its value is - as
Therefore the solution of the given differential equation is
-
Find the general solution of each of the following differential equations:
can be written as
Integrating on both the sides,
formula:
formula:
Find the general solution of each of the following differential equations:
Given:
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Find the general solution of each of the following differential equations:
Given:
⇒
⇒
⇒
⇒
Find the general solution of each of the following differential equations:
Given:
⇒
⇒
⇒
Find the general solution of each of the following differential equations:
cos x(1 + cos y)dx – sin y(1 + sin x)dy = 0
Given: cosx(1+cosy)dx-siny(1+sinx)dy=0
Dividing the whole equation by (1+sinx)(1+cosy), we get,
⇒
⇒ log|1+sinx|+log|1+cosy|=logc
⇒ (1+sinx)(1+cosy)=c
Find the general solution of each of the following differential equations:
sin3 x dx – sin y dy = 0
We have,
⇒
⇒
⇒
⇒
⇒
Find the general solution of each of the following differential equations:
⇒
⇒ (Using sin(A+B)-sin(A-B)=2sinBcosA)
⇒
⇒
⇒
⇒ sinx+log|cosecy-coty|+c=0
Find the general solution of each of the following differential equations:
Given:
⇒
⇒ (Using, 2cos2a=1+cos2a)
⇒
⇒
⇒
Find the general solution of each of the following differential equations:
Here we have,
⇒
Taking cosx as t we have,
⇒
⇒
So we have,
⇒
⇒
⇒
Find the particular solution of the differential equation given that y = 0 when x = 1.
Given:
⇒
⇒
⇒
We have,
⇒
⇒
⇒
Find the particular solution of the differential equation x(1 + y2) dx
-y(1 + x2) dy = 0, given that y = 1 when x = 0.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Find the particular solution of the differential equation given that y = 0 when x = 0.
⇒
⇒
⇒
⇒
⇒
⇒ For , we have
⇒
⇒
⇒
⇒
Solve the differential equation (x2 – yx2)
dy + (y2 + x2y2) dx = 0, given that y = 1 when x = 1.
⇒
For y=1,x=1, we have,
⇒
⇒ 1
Hence, the required solution is:
⇒
Find the particular solution of the differential equation given that y = 1 when x = 0.
Given: Separating the variables we get,
⇒
⇒ Substituting , we have,
⇒
For y=1 and x=0, we have,
⇒
⇒
⇒ Hence, the particular solution will be:-
⇒
Find the particular solution of the differential equation given that when x = 1.
Given:
⇒
Let Then,
And
We have,
⇒
For we have,
⇒
Solve the differential equation given that y(0) = 1.
We have,
⇒
⇒
For y=1, x=0, we have,
c =
⇒
⇒
Thus,
The particular solution is:
Solve the differential equation given that y(2) = 3.
Given:
⇒
⇒
⇒
For x=2 and y=3, we have,
c = 3log3 – 4
Hence, the particular solution is,
⇒ y(x + 1)2 = 27
Solve given that y = 0 when x = 2.
we have, , Integrating we get,
,
given that y=0 when x=2
⇒
now putting x=2 and y=0,
⇒
⇒
Thus, the solution is:
Solve given that y = 1 when x = 0.
we have,
Given that: y=1 when x=0,
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
For y=1, when x=0, we have,
⇒
Solve given that y = 1 when x = 0.
we have,
given that: y=1 when x=0
⇒
⇒
⇒
⇒
⇒ is the particular solution…
Solve given that y = 2 when x = 0.
we have:
Given that, y=2 when x=0
⇒
⇒ …integrating both sides
⇒
⇒
⇒
⇒ …is the particular solution
Solve given that y = 2 when
we have
Given that, y=2 when x=
⇒
⇒
⇒
⇒
⇒ +c
⇒ Thus,
The particular solution is :-
Solve (1 + x2) sec2 y dy + 2x tan y dx = 0, given that when x = 1.
we have, (1 + x2) sec2 y dy + 2x tan y dx = 0,
Given that, when x=1
⇒
⇒
⇒
⇒
For
We have,
c = 2,
Hence the required particular solution is:-
Find the equation of the curve passing through the point whose differential equation is sin x cos y dx + cos x sin y dy = 0.
we have, sin x cos y dx + cos x sin y dy = 0
⇒
⇒
⇒
⇒
Given that, coordinates of point
⇒
⇒
…is the required particular solution
Find the equation of a curve which passes through the origin and whose differential equation is
Given,
⇒
Let
⇒
⇒
⇒
⇒
⇒
⇒
For the curve passes through (0,0)
We have, c =
A curve passes through the point (0, -2) and at any point (x, y) of the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. Find the equation of the curve.
Given that the product of slope of tangent and y coordinate equals the x-coordinate i.e.,
We have,
⇒
⇒
For the curve passes through (0, -2), we get c = 2,
Thus, the required particular solution is:-
A curve passes through the point (-1, 1) and at any point (x, y) of the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve.
Given :
⇒
⇒
⇒
The curve passes through (-2, 1)we have,
c = 0,
In a bank, principal increases at the rate fo r% per annum. Find the value of r if ` 100 double itself in 10 years.
(Given loge 2 = 0.6931)
Given:
Here, p is the principal, r is the rate of interest per annum and t is the time in years.
Solving the differential equation we get,
⇒
⇒
⇒
As it is given that the principal doubles itself in 10 years, so
Let the initial interest be p1 (for t = 0 ), after 10 years p1 becomes 2p1.
Thus, for ( t = 0) …(i)
…(ii)
Substituting (i) in (ii), we get,
⇒
⇒
⇒
⇒
⇒
Rate of interest = 6.931
In a bank, principal increases at the rate of 5% per annum. An amount of ` 1000 is deposited in the bank. How much will it worth after 10 years?
(Given e0.5 = 1.648)
Given: rate of interest = 5%
P(initial) = Rs 1000
And,
⇒
⇒
⇒
For t = 0, we have p = 1000
For t = 10 years we have,
Thus, principal is Rs1648 for t = 10 years.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Given:
Volume V
⇒ (constant)
⇒
⇒
⇒
For t = 0, r = 3 and for t = 3, r = 6, So, we have,
⇒
⇒
⇒
So after t seconds the radius of the balloon will be,
⇒
⇒
⇒
⇒
⇒
Hence, radius of the balloon as a function of time is
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?
Let y be the bacteria count, then, we have,
rate of growth of bacteria is proportional to the number present
So,
Where c is a constant,
Then, solving the equation we have,
Where k is constant of integration
And we have for t = 0, y = 10000,
…(i)
For t = 2hrs, y is increased by 10% i. e. y = 110000
⇒ from (i)
⇒
⇒
⇒
When y = 200000, we have,
⇒
⇒
⇒
⇒
Hence, t =