mixture: variable separable differential equations type

Showing posts with label variable separable differential equations type. Show all posts

Wednesday, June 16, 2021

solve y log y dx – x dy = 0

ncert cbse mathematics class 12th differential equations

exercise 9.4 variable separable differential equations type

7. solve y log y dx – x dy = 0

This is a variable separable differential equation

separating the variables [dividing each term with y.log y . x

[dx/x ] - [ dy / (y.log y ) ] = 0

integrating using integration formula

and rearranging [ dy / (y.log y ) ] =[ (1/y)dy / logy ] and noting

that integral of {(f ') / f }form is ln(f)

log x - log{logy}+logC = 0

here log stands for ln (natural logarithm)

using properties of log

log[ Cx / logy] = 0

Cx / log y = e^0

Cx / logy =1

log y = Cx

or y = e^(Cx)

8. solve (x^5) {dy/dx} = − (y^5 )

This is a variable separable differential equation

separating the variables

dy/(y^5 ) = -dx /(x^5 )

(y^(-5) )dy =- (x^(-5) )dx

integrating using integration formula

[(y^(-4)]/(-4) = -[(x^(-4)]/(-4) +C/(-4)

so that

[(x^(-4)]+[(y^(-4)] =C

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ncert cbse 12th mathematics

exercise 9.4

differential equations

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

solution

5.solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution

7. solve y log y dx – x dy = 0

solution

8. solve (x^5) {dy/dx} = − (y^5 )

solution

solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

ncert cbse mathematics class 12th differential equations

exercise 9.4 variable separable differential equations type

5.solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

This is a differential equation of the type variable separable.

( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

dy ={ [ e^x – e^(–x) ] / [e^x + e^( –x)] } dx

now integrate both sides

keeping in mind that integral of {(f ') / f }form is ln(f)

{ here ln refers to natural logarithms }

y=ln{[e^x + e^( –x)]} +C

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

This is a differential equation of the type variable separable.

dy / [ 1 + (y^2) ] = [ 1+( x^2) ]dx

now integrating on both sides using integration formula

arc[tan y] =x +[(x^3)/3] +C

note that arc[tan y] stands for inverse tan of (y)

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ncert cbse 12th mathematics

exercise 9.4

differential equations

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

solution

5.solve ( e^x + e^( –x) ) dy – ( e^x – e^(–x) ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution

Monday, June 14, 2021

solve (dy/dx)+y=1

ncert cbse mathematics class 12th differential equations

exercise 9.4 variable separable differential equations type

3. solve (dy/dx)+y=1

(dy/dx)+y=1

(dy/dx)=1-y

separating the variables

dy / [1-y] =dx

integrating,

[remember to divide by (-1), the coefficient of y]

{ln[1-y] }/(-1)=x+ln C

{ here ln refers to natural logarithms }

-ln(1-y) =x+ln C

lnC +ln(1-y) = -x

using properties of logarithms

ln [C(1-y)]= (-x)

getting rid of logarathmic funtion

C(1-y)= e^(-x)

(1-y) = (1/C) e^(-x)

y=1-(1/C) e^(-x)

put (1/C) =(-A)

y = 1 + A e^(-x)

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

divide each term by tan y tan x

[ { (sec x)^2}/ tanx ]dx + [ { (sec y)^2}/ tan y ]dy = 0

integrating term by term

keeping in mind that integral of {(f ') / f }form is ln(f)

{ here ln refers to natural logarithms }

ln(tan x) +ln(tan y) =lnC

using properties of logarithms

ln(tanx tany ) =lnC

tan x tan y = C

=================================================

ncert cbse 12th mathematics

exercise 9.4

differential equations

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

solution

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Sunday, June 13, 2021

solve dy/dx = [1-cosx]/[1+cosx]

ncert cbse mathematics class 12th differential equations

exercise 9.4 variable separable differential equations type

solve dy/dx = [1-cosx]/[1+cosx]

here this is of the type variable separable

dy/dx = [1-cosx]/[1+cosx]

dy = { [1-cosx]/[1+cosx] }dx

using trigonometric formula

1-cosx =2 [ sin(x/2)] ^2

1+cosx =2 [ cos(x/2)] ^2

dy = { [{2 [ sin(x/2)] ^2} ] / {2 [ sin(x/2)] ^2} }dx

dy =[tan(x/2)] ^2 dx

we again have to use trigonometric formula to change the

[tan(x/2)] ^2 in terms of [sec(x/2)] ^2 -1 before integrating

dy ={[sec(x/2)] ^2 -1 }dx

now integrating on both sides using integration formula

y = { [tan(x/2)] /[1/2] } -x + C

do not forget to divide by the coefficient of x while integrating

y=2 [tan(x/2)] - x +C

you can use the following for more explanation

2. solve (dy/dx) = sqrt[4-(y^2)]

separating the variables

dy/{sqrt[4-(y^2)]} =dx

now integrating on both sides using integration formula

and noting that 4=(2^2) so that a=2

we get

arc[sin(y/2)]=x+C

y/2 =sin(x+C)

y=2 sin(x+C)

=================================================

ncert cbse 12th mathematics

exercise 9.4

differential equations

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

2. In each of the following find the value of ‘k’, for which the points are collinear.

(7, –2), (5, 1), (3, k)

solution

(ii) (8, 1), (k, – 4), (2, –5)

solution

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order

solution

9. Find the coordinates of the points which divide the line segment joining

A(– 2, 2) and B(2, 8) into four equal parts.

solution

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution

4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

solution

2. Find the coordinates of the points of trisection of the line segment joining

(4, –1) and (-2,-3)

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3

solution

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

solution

(ii) (– 5, 7), (– 1, 3)