Solve the equation 2 cosh 2x - sinh 2x = 2?
2 cosh 2x - sinh 2x = 2 2 cosh²x + 2 sinh²x - 2 sinhx coshx = 2 2 sin²x + 2 sinh²x - 2 sinhx coshx = 0 sinx(2sinhx - cosx) = 0 Answer: x1 = 0 x2 = artanh(1/2) = 1/2 ln(3)
Solve cosh^-1(2x)=sinh^-1x?
cosh^-1(2x) = sinh^-1(x) recall that cosh^-1(x) = ln [ x + √(x^2 - 1) ] and sinh^-1(x) = ln [ x + √(x^2+1) ] so ln [2x + √(4x^2 - 1) ] = ln [ x + √(x^2+1) ] 2x + √(4x^2 - 1) = x + √(x^2 + 1) √(x^2 + 1) - √(4x^2 - 1) = x squaring x^2 + 1 + 4x^2 - 1 - 2√(x^2 + 1)√(4x^2 - 1) = x^2 2√(x^2 + 1)√(4x^2 - 1) = 4x^2 => √(x^2 + 1)√(4x^2 - 1) = 2x^2 again squaring => (x^2 + 1)(4x^2 - 1) = 4x^4 4x^4 + 3x^2 - 1 = 4x^4 3x^2 - 1 = 0 3x^2 = 1 x^2 = 1/3 x = 1/√3 (ignoring invalid negative value) x = (1/3)√3
How do I solve [math]2 \cosh 2 x + 10 \sinh 2 x = 5[/math]?
The hyperbolic functions sinh x and cosh x are defined asSinh x = (e^x - e^-x)/2 and cosh x = (e^x + e^-x)/2Hence, the given equation is2 {(e^2x + e^-2x)/2} + 10 { (e^2x-e^-2x)/2} = 5e^2x + e^-2x +5 (e^2x - e^-2x) = 5Simplifying we get 6e^2x - 4 e^-2x -5 =0Dividing by e^-2x we get6e^4x -5e^2x -4 = 0(3e^2x -4)(2e^2x +1) = 0e^2x =4/3 or e^2x = -1/2e^x cannot be negative.So e^2x =4/3Taking log , 2x = log 4/3 = log 4 - log 3x = log 2 - 1/2(log 3)
Solve the equation 2 cosh 2x - sinh 2x = 2 using exponential function?
If this --> 2 cosh ( 2x) - sinh (2x) = 2 ---> is the question, then the answer is x = [ log 3 ] / 2.
Convert the parametric equations x = 3cosh t y = 2 sinh t to rectangular form?
Rearranging. cosh t = x/3 and sinh t = y/2 Use the hyperbolic identity cosh² t - sinh² t = 1 Then (x/3)² - (y/2)² = 1 x² / 9 - y² / 4 = 1 This is the equation of a hyperbola.
Solve 3 cosh 2x + 5 cosh x= 22?
use the double-angle identity cosh 2x = 2 cosh² x - 1 3 cosh 2x + 5 cosh x = 22 6 cosh² x - 3 + 5 cosh x = 22 6 cosh² x + 5 cosh x - 25 = 0 (2 cosh x + 5) (3 cosh x - 5) = 0 Only the positive root cosh x = 5/3 gives a real answer for x. x = arcosh 5/3
How do I solve the differentiation of x^-cos^2x=y.x^sin^x? What is dy/dx=?
Clearly state question next time. It becomes difficult to understand the question itself.This answer is based on how I understood the question. Thank you.
What is the solution of the differential equation: Y"-Y= 2Cosh X?
First solve the homogeneous equation[math]y''-y=0\rightarrow y=A\cosh x+B\sinh y[/math]Now look for a particular solution. Since the non-homogeneous part contains [math]\cosh x[/math], the particular solution should have the form [math]C\cosh x+D\sinh x[/math]. However, since [math]\cosh x[/math] is already part of the homogeneous solution, then the particular solution is of the form [math]Cx\cosh x+Dx\sinh x[/math]. Setting this particular solution in the ODE yields[math]C\left[x\cosh x+2\sinh x\right]+D\left[x\sinh x+2\cosh x\right]-Cx\cosh x-Dx\sinh x=2\cosh x[/math]Rearranging[math]2C\sinh x+2D\cosh x=2\cosh x[/math]Therefore [math]C=0[/math] and [math]D=1[/math]Finally, the general solution is[math]y=A\cosh x+B\sinh+x\sinh x[/math]
Express cosh 2x and sinh 2x in exponential form; solve for real values of x for the following:?
2 cosh(2x) - sinh(2x) = 2 ==> 2 * (1/2)(e^(2x) + e^(-2x)) - (1/2)(e^(2x) - e(-2x)) = 2, by definition of sinh and cosh ==> 2(e^(2x) + e^(-2x)) - (e^(2x) - e(-2x)) = 2 * 2, multiplying both sides by 2. Simplifying: e^(2x) + 3 e^(-2x) = 4 ==> e^(4x) + 3 = 4e^(2x), multiplying both sides by e^(2x) ==> e^(4x) - 4e^(2x) + 3 = 0 Solve by factoring: (e^(2x) - 3)(e^(2x) - 1) = 0. ==> e^(2x) = 3 or 1 ==> 2x = ln 3 or 0 ==> x = (1/2) ln 3 or 0. I hope this helps!