solving a second order differential equation using reduction of order method if one of the solutions is given

solve y"-y = 0 if y=coshx is one of the solutions
using the formula for reduction of order
 substitute p=0 since there is no term in y' and
 use y1 = cosh x the given solution simplify and then integrate
 integration formulae

 
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problem on scalar triple product stp of vectors

show that [axb bxc cxa] = [a b c]^2
where a,b,c are vectors

use the formula for crossproduct of four vectors


then use the property that if a vector is repeated  in an stp, then the value of the stp is zero

then use the property that interchange of vectors in an stp changes the sign of the stp.

























some questions in mathematics for the tamilnadu higher secondary 12th exam, isc ,  puc pre degree etc.
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to find the equation of the tangents that can be drawn from a given point to a given hyperbola in standard form

find the equation of the tangents from (1,2) to the hyperbola 2(x^2) -3(y^2) =6

 divide by the term on the rhs to reduce the given equation to standard form
extract the value of a and b
use the equation of the tangent to the hyperbola in standard form and use the values of a and b
use the condition that the tangent passes through the given point to form an equation in m and solve for m
and resubstitute

 
 

 
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solving a fourth degree equation with real coefficients if you are given one of the imaginary roots

solving a fourth degree equation with real coefficients if you are given one of the imaginary roots

solve x^4 - 4 x^3  + 11 x^2  -14x +10 =0  given that  1+2i   is  one of the roots.

since the coefficients of the given equation are real, the complex roots will occur in conjugate pairs

so 1-2i is another root.

using this you can find a quadratic factor of the given fourth degree expression.

use long division or comparision of coeffecients to extract the other quadratic factor and solve it

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checking consistency of equations using determinants

Solve using determinants (Cramers rule)

2x + 3y = 8

 4x + 6y = 16

n = 2 unknowns
r =1

n-r = 2-1 = 1

therefore one unknown can be chosen arbitrarily

put y = k  (arbitrary ) in the first equation  2x +3y =8

2x + 3k = 8

solving x = (8 - 3k) / 2


so solution is

x = (8 - 3k) / 2

y = k

where k is any number.














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problem on trigonometry

start from the LHS
use the formula for cos2x and sin2x

factorise the numerator
adjust the 1 in the denominator using trigonometry identity
write as a square

divide with sinx to change to cot x


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integral of arc (sin x)

integral of arc (sin x) using integration by parts

change to arc (sin x) . 1 and use integration by parts with

u = arc(sin x)  and v =1

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

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