Oasis Academy Shirley Park



Y13 Further Maths – Spring 1 Knowledge OrganiserChapter 6 – Hyperbolic FunctionsKEY DEFINITIONSHyperbolic functions and their inverses are defined in terms of the constant e:sinhx=ex-e-x2, ?x∈R arsinh?x=lnx+x2+1coshx=ex+e-x2, x∈R arcosh?x=lnx+x2-1,???x≥1tanh=sinhxcoshx=e2x-1e2x+1, x∈R artanh?x=12ln1+x1-x,? ???x<1ODD AND EVEN FUNCTIONSLike sin?(x), sinh?(x) is an odd functionFor all values of x: sinh-x=-sinh?(x)The graph of y=sinhx has rotational symmetry, order 2, around (0,0)Like cos?(x), cosh?(x) is an even functionFor all values of x: cosh-x=cosh?(x)The graph of y=coshx has reflective symmetry in the y-axis.GRAPHS OF THE 6 FUNCTIONS (& THEIR INVERSE FUNCTIONS)TRIGONOMETRIC IDENTITIESTo convert standard trigonometric identities, into hyperbolic identities, follow Osborn’s Rule: Replace sin→sinh and cos→coshNegate any explicit or implied product of two sines.Therefore:cosh2x-sinh2x≡1sinh?(A±B)≡sinhAcosh?(B)±coshAsinh?(B)cosh?(A±B)≡coshAcosh?(B)±sinhAsinh?(B)DERIVATIVES AND INTEGRALS (THAT ARE NOT IN THE FORMULA BOOKLET)ddxsinh?(f(x))=f'(x)cosh?(f(x)) ddxcosh?(f(x))=f'(x)sinh?(f(x))ddxtanh?(f(x))=f'(x)sech?(f(x)) ddxartanh?(x)=11-x2tanh?(x)=lncoshx+cChapters 7 & 8 – Differential EquationsA differential equation is defined as any equation which involves a derivative function.It has a general solution (another equation, which describes the same pairs of x and y, but does not involve any derivative functions).The general solution will involve a ‘+c’, which could be any real number. This creates an infinite family of curves which form the ‘general solution’.It has a particular solution, the particular curve from the family of curves which form the general solution (this particular solution depends on the boundary conditions of the problem)Method 1: Separating the Variables (1st order ODES)Some 1st order ODEs can be solved using the “separating the variables” method.They must be of the form dydx=fxgy, so that we can manipulate the expression into 1gydy=fx?dx and simply integrate our way to the general solution.Method 2: Multiplying by the Integrating Factor (1st order ODEs)1st order ODEs of the form dydx+Pxy=Qx, can be solved by multiplying all 3 terms by the integrating factor eP(x)dx, and then using the reverse product rule.*Note: You may have to manipulate a 1st order ODE to get it into one of the above 2 forms.Method 3: Homogeneous 2nd order ODEs3773170252095They’re the same a, b, c as the co-efficients of the original question. 00They’re the same a, b, c as the co-efficients of the original question. 2nd order homogenous ODEs that you will be solving will always be of the form:ad2ydx2+bdydx+cy=0Step 1: Form the auxiliary equation am2+bm+c=0Solve the quadratic to find the solutions to the auxiliary (we’ll call the solutions α and β) Step 2: Use the classification of the solutions to create a complementary function:When the auxiliary equation has two real distinct roots α and β, the complementary function is y=Aeαx+Beβx, where A and B are arbitrary constants.When the auxiliary equation has two equal roots α, the complementary function isy=A+BxeαxIf the auxiliary equation has two imaginary roots ±iω, the complementary function is y=Acosωx+Bsinωx where A and B are arbitrary constants.If the auxiliary equation has two complex roots p±iq, the complementary function is y=epxAcosqx+Bsinqx*In the homogeneous case, the general solution is equal to the complementary function.*We can now plug in boundary conditions (if provided) to find the particular solution.Method 4: Non-homogeneous 2nd order ODEs4319905192405They’re called non-homogenous because the RHS ≠000They’re called non-homogenous because the RHS ≠02nd order homogenous ODEs that you will be solving will always be of the form:ad2ydx2+bdydx+cy=f(x)*Method 4 begins by finding the complementary function, exactly as laid out above.Step 3: Select a particular integral by examining the form of the RHS f(x) – [see table!]Differentiate twice to find dydx and d2ydx2Substitute these expressions into the original functions LHSCompare co-efficients to find values for λ and μImportant Note: If the PI is a part of your CF (identically the same), then you must multiply your PI by a “cheeky little x”Form of f(x)Form of your particular integralkλax+bλ+μxax2+bx+cλ+μx+νx2kepxλepxmcosωxλcosωx+μsinωxmsinωxλcosωx+μsinωxmcosωx+nsinωxλcosωx+μsinωxStep 4: Form your general solution: General Solution= Complementary Function+Particular IntegralStep 5 (optional): Substitute your boundary conditions into your general solution in order to find your particular solution. ................
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