mixture: 2017

Tuesday, March 14, 2017

integral of [e^2x]cos3x using integration by parts

integral of [e^2x]cos3x using integration by parts

Take the integral as I

take cos3x as the first function and apply integration by parts , simplify and then apply integration by parts to the integral obtained in the first answer.

Simplify to get the original integral I in the right hand side and change it to I

Solve for I from this equation

formulae on integration

PAGE 1 BASIC INTEGRATION

PAGE 2 INTEGRATION BY SUBSTITUTION

PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR

PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

Sunday, February 5, 2017

linear differential equation y dx + [x-(y^2)]dy = 0

linear differential equation from ncert cbse

solve y dx + [x-(y^2)]dy = 0

The equation contains only one x term

Therefore try to solve for [dx/dy] and compare with the linear differential equation of the form

[dx/dy] + Px = Q

Then compare and get the values of P and Q

find the integrating factor of the linear differential equation using the formula

I.F. = e^[integral of P dx]

use the property e^[ln[f(y)] = f(y)

then use the solution

x[I.F.] = integral of [ Q * I.F.]dy +C

Variable separable differential equation

*find the equation of the curve passing through (0,pi/4) whose differential equation is
sinx cosy dx + cosx siny dy =0
solution of variable separable differential equation find the equation of the curve passing through (0,pi/4) whose differential equation issinx cosy dx + cosx siny dy =0

* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

solution of variable separable differential equation from ncert cbse miscellaneous problem

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

*linear differential equation

solve [ { e^[-2sqrt(x)] / sqrt(x) } - {y / sqrt(x)}] [dx/dy] = 1 [x is not 0]
solution of linear differential equation of ncert cbse miscellaneous differential equation

*linear differential equation of the type
solve y dx + [x-(y^2)]dy = 0
solution of linear differential equation of the type [dx/dy] + Px = Q

*find a particular solution of the differential equation [dy/dx]+ycotx =4xcosecx [x is not 0] given that y =0 if x=pi/2
solution of linear differential equation from ncert cbse miscellaneous [dy/dx]+ycotx =4xcosecx [x is not 0] given that y =0 if x=pi/2

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to find the orthogonal trajectory of y = (k/x)

formulae on integration

PAGE 1 BASIC INTEGRATION

PAGE 2 INTEGRATION BY SUBSTITUTION

PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR

PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work

Wednesday, February 1, 2017

linear differential equation

linear differential equation from ncert cbse miscellaneous differential equation

find a particular solution of the differential equation [dy/dx]+ycotx =4xcosecx [x is not 0] given that y =0 if x=pi/2

the equation is an example of linear differential equation, so compare [dy/dx]+ycotx =4xcosecx with the standard linear differential equation [dy/dx]+Pycotx =Q

Identify the values of P and Q

here P=cotx

Q=4xcosecx

Find the integrating factor[I.F.] = e^[integral of P ]

use the result e^[ln{f(x)}] = f(x)

use the solution

y[I.F.] = integral of [ Q* I.F. ] +C

Variable separable differential equation

solution of variable separable differential equation from ncert cbse miscellaneous problem

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

*linear differential equation

solve [ { e^[-2sqrt(x)] / sqrt(x) } - {y / sqrt(x)}] [dx/dy] = 1 [x is not 0]
solution of linear differential equation of ncert cbse miscellaneous differential equation

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Tuesday, January 31, 2017

linear differential equation for cbse ncert miscellaneous problem

linear differential equation for cbse ncert miscellaneous problem

solve [ { e^[-2sqrt(x)] / sqrt(x) } - {y / sqrt(x)}] [dx/dy] = 1 [x is not 0]

here there is only one term in y

so we have to rearrange the equation in the form

[dy/dx] + Py = Q

identify the values of P and Q

find the integrating factor e^[integral of P]

and use in the solution

y[integrating factor] = integral of [Q*integrating factor] + C

Variable separable differential equation

solution of variable separable differential equation from ncert cbse miscellaneous problem

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

non homogeneous equation with the substitution x=vy

miscellaneous problem from ncert differential equation where the substitution x=vy still works even though is not a homogeneous equation

solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]

here there are terms containing [x/y]
so extract dx/dy call the value of dx/dy as f[x,y]
replace x with tx and y with ty and check if the function is homogeneous

here f[tx,ty] is not equal to f[x,y]
therefore the function is not homogeneous

but the substitution v =(x/y) will still work for this problem

Variable separable differential equation

* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

solution of variable separable differential equation from ncert cbse miscellaneous problem

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

miscellaneous problem from differential equations

miscellaneous problem from differential equations
find the equation of the curve passing through (0,pi/4) whose differential equation is
sinx cosy dx + cosx siny dy =0

here cosy with dx and cosx with dy should be removed.

So divide each term with cosy cosx

the resulting equation is of variable separable type.

integrate term by term

Now to get the value of C, use the given condition that the curve passes through (0,pi/4)

put x =0 and y = pi/4 and solve for C.

Variable separable differential equation

* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

solution of variable separable differential equation from ncert cbse miscellaneous problem

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

variable separable differential equation for cbse ncert 12th

variable separable differential equation for cbse ncert 12th

show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

first separate the variables , then, complete the squares in each term, integrate term by term
then use the formula for arctanx+arctany

homogeneous differential equation
prove that [ x^2 - y^2 ]=c [ x^2 - y^2 ]^2 is a solution of
[ x^3 - 3x (y^2) ] dx = [ y^3 - 3x^2y ] dy

solution of homogeneous differential equation from miscellaneous problems of ncert cbse 12th mathematics

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Thursday, January 26, 2017

solution of homogeneous differential equation

solution of homogeneous differential equation

prove that [ x^2 - y^2 ]=c [ x^2 - y^2 ]^2 is a solution of
[ x^3 - 3x (y^2) ] dx = [ y^3 - 3x^2y ] dy

first solve for (dy/dx)
take the function as f(x,y)
now check if the equation is a homogeneous differential equation
replace x and y with tx and ty
simplify and see if the t can be cancelled off to get back f(x,y)
If the equation is a homogeneous differential equation, put y=Vx
(dy/dx) = V + x(dV/dx)
simplify and separate the variables and then integrate

Variable separable differential equation

solution of variable separable differential equation from ncert cbse miscellaneous problem

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

*linear differential equation

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

form the differential equation of the family of circles in the first quadrant which touch the coordinate axes

Take the radius as r so that the centre is (r,r)

write the equation of the circle in terms of r

r is the aribitrary constant.

so differentiate it one time with respect to x and use

that to solve for r

substitute into (x-r) and (y-r) and substitute into the equation of the

circle to eliminate r

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Tuesday, January 24, 2017

integral of e^(2-3x) from 0 to 1 using limit of sums

integral of e^(2-3x) with limit 0 to 1 using limit of sums

identify a =0 , b=1

nh = b-a = 1

f(x) = e^(2-3x)

find f(a) = f(0)
f(a+h)=f(h)
f(a+2h) = f[2h]
till the pattern can be identified,
f(a + (n-1) h ) = f[(n-1) h] etc

simplify using properties of sum of n terms of a GP.
and use the limit [(e^h) -1] / h tends to 1 as h tends to 0

integral of 2*cube of (tanx) with 0 to pi/4 as limits

integral of 2* (tanx)^3 with 0 to pi/4 as limits

write (tanx)^3 as the product of tanx and (tanx)^2

use trigonometric formulae to rewrite (tanx)^2 as [ (secx)^2 -1]

multiply the tanx into [ (secx)^2 -1] and split into two integrals

for the first one, use the substitution t =tanx

the second integral is a direct formula.

Friday, January 20, 2017

integral of { (cosx)^2 / [ (cosx)^2+ 4 (sinx)^2 ] }

integral of { (cos x)^2 / [ (cos x)^2+ 4 (sin x)^2 ] }

use trigonometric formulae to change (sin x)^2 = 1 - (cos x)^2 so that the intergral is completely in terms of (cos x)^2 .
Now try to write the numerator in terms of the denominator.
introduce a (-3 ) in the numerator and denominator and add and subtract 4

split it into two terms and then two integrals

The second integral contains (cos x)^2 .

divide each term with (cos x)^2 to get (sec x)^2

use trigonometric formulae tochange (sec x)^2 = 1+ (tan x)^2 in the denominator only

use substitution t = tan x and change the limits.

Thursday, January 19, 2017

integral using substitution and then integration by parts

integral using substitution and then integration by parts

integral of [1 / (x^4)][sqrt( 1 + (x^2))][ log( 1 + (x^2)) - 2log( x)]

first simplify using property of logarithms

take x^2 common from the sqrt term obtain [1+1/(x^2)] and try to get the same term inside the log expression

cancel off the x to get 1 /[x^3] then use substitution

then use integration by parts with log(t) as the first function

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Tuesday, March 14, 2017

Sunday, February 5, 2017

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Wednesday, February 1, 2017

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Tuesday, January 31, 2017

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Thursday, January 26, 2017

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

Tuesday, January 24, 2017

Friday, January 20, 2017

Thursday, January 19, 2017

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