√1000以上 y'=y/x cos(y/x) 550145-Xcos(y/x)dy/dx=y cos(y/x)+x
Sep 02, 09 · use implicit differentiation, and expand the cos(xy) term using the cos addition rule xcos(xy) = x cosx cosy sin x sin y =0 take d/dx of each term d/dx (x) =1 d/dx(cosx cosy) = sinx cosy cosx siny dy/dx d/dx(sinx sin y) = cosx siny sin x cos y dy/dx so you have 1sinx cosy cosx siny dy/dx cosxsiny sin x cosy dy/dx =0Answer to Find the second derivative, y"(x), of 3y2 = 2x3 x cos y y"(x) = 4( cos y 3)(y')2/ 3y sin y y"(x)= 2x2 Sin X At Which The Tangent Line Is Horizontal (answers As A Commaseparated List Use N To Represent Any Integer)
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Xcos(y/x)dy/dx=y cos(y/x)+x
Xcos(y/x)dy/dx=y cos(y/x)+x-Y(t) = c1 cos(t) c2 sin(t) −cos(t)ln(sec(t) tan(t)) #17 Verify that the given function y1 and y2 satisfy the corresponding homogeneous equation;Oct 28, 11 · 1/y * y' = (6 sinx) * lnx (1/x) * (6 cosx) > simplify & expand 1/y * y' = 6 ( sinx lnx cosx / x ) > multiply both sides by y to isolate y' y' = 6 y ( sinx lnx cosx/x) > substitute y back in from the original equation
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Y = Cos(x2) Find Y' AND Y'' 2 Find An Equation Of The Tangent Line To The Curve At The Given Point Y = (1 4x)12, (0, 1) 3 Find The Xcoordinates Of All Points On The Curve F(x) = Sin 2x ?Solve the initial value equation y = x cos x 2 sin x with y(0) = 1 and y (0) = 2 Select one O a y = x cos x x 1 O b y = x sin x 1 O c y = cos x x O d y = x cos x 1 Consider the system X' = AX where is a 3 x 3 matrix with a complex eigenvalue a fi corresponding to an eigenvector a blFree math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor
Mar 31, 08 · Let's step through it y′′ y=sin(x)xcos(x) y ″ y = s i n ( x) x c o s ( x) The auxiliary equation is m21 =0, m= i, m = −1 m 2 1 = 0, m = i, m = − 1 Since the solutions to the auxiliary equation are complex, we have yc = C1cos(x)C2sin(x) y c = C 1 c o s ( x) C 2 s i n ( x)Cov(X,Y) = E((XE(X)) * (YE(Y)) ) (which happens to be equal to E(XY)E(X)E(Y) the definition you may have seen) But in any case, from the definition you can check Cov(XZ,Y) = Cov(X,Y) Cov(Z,Y) and that's why you can 'expand brackets' (and similarly in the second 'slot')Hint Use the Method of undetermined coefficients Since x is a polynomial of degree one and i is not a solution of the characteristic polynomial, you should try with y p ( x) = ( A x B) cos ( x) ( C x D) sin ( x) where the constants A, B, C, D have to be determined by plugging y p in the differential equation
Graph y=cos(x) Use the form to find the variables used to find the amplitude, period, phase shift, and vertical shift Find the amplitude Amplitude Find the period of Tap for more steps The period of the function can be calculated using Replace with in the formula for periodFind dy/dx e^ycos(x)=1sin(xy) Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate using the Product Rule which states that is where and The derivative of with respect to is Simplify the expressionJun 03, 16 · cos(xy) = cos\ x* cos\ y sin\ x* sin\ y cos(xy) = cos\ x*cos\ y sin \ x*sin\ y sin^2 x cos^2\ x= 1 cos(xy) = cos\ x* cos\ y sin\ x* sin\ y cos(xy) = cos\ x
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Solution for 12) dy dx = x cos (6x²), y(0) = 7 A) y == sin (u) B) y = sin(6x2) 12 C) y = sin (6x²) D) y sin(6x²) 2 ='Apr 07, 17 · I'm going to assume you meant #sin(xy)/(sinxcosy)=1cotxtany# and just forgot parentheses on the left side denominator and the plus on the right side, otherwise the identity is false Let's start working the left hand side by using the sine addition formula #sin(alphabeta)=sinalphacosbetacosalphasinbeta#Answer to Sketch the region enclosed by the curves and find its area y = \cos \pi x, y = 4x^2 1 By signing up, you'll get thousands of
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge andSince the general solution to the homogeneous equation, y'' y= 0 are \sin(x) and \cos(x), we look for a solution of the form y(x)= u(x)\cos(x) v(x)\sin(x) As I pointed out before, cos 2 ( x ) = cos ( 2 xWe have mathy=/mathmathcos(xy)/math Differentiating, math\dfrac{dy}{dx} = \sin(xy)(xy)'/math Chain rule math\Rightarrow \dfrac{dy}{dx} = \sin(xy
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Sin(x y) = sinxcosy cosxsiny sin(x y) = sinxcosy cosxsiny cos(x y) = cosxcosy sinxsiny cos(x y) = cosxcosy sinxsiny tan(x y) = tanxtany 1 tanxtany tan(x y) = tanx tany 1tanxtany HalfAngle Formulas sin 2 = q 1 cos 2 cos 2 = q 1cos 2 tan 2 = q 1cos tan 2 = 1 cosx sinx tan 2 = sin 1cos DoubleAngle Formulas sin2 = 2sin cos cos2 = cos2 sin2 tan2 = 2tan 1 tan2Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyIf units of degrees are intended, the degree sign must be explicitly shown (eg, sin x°, cos x°, etc) Using this standard notation, the argument x for the trigonometric functions satisfies the relationship x = (180x/π)°, so that, for example, sin π = sin 180° when we take x = π In this way, the degree symbol can be regarded as a
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Here is a "reductio in absurdum" approach y ′ ycos(x) = sin(x)cos(x) y ′ = cos(x)(sin(x) − y) now set sin(x) − y = z, we get cos(x) − y ′ = z ′ cos(x) − z ′ = cos(x)z z ′ = cos(x)(1 − z) now let 1 − z = a, − z ′ = a ′, − a ′ = cos(x)aJan 15, 17 · dy/dx = x^cosx(sinxlnx cosx/x) y = x^cosx Take the natural logarithm of both sides lny = ln(x^cosx) Use the logarithm law for powers, which states that loga^n = nloga lny = cosxlnx Use the product rule to differentiate the right hand side d/dx(cosx) = sinx and d/dx(lnx) 1/y(dy/dx) = sinx(lnx) cosx(1/x) 1/y(dy/dx) = sinxlnx cosx/x dy/dx = (sinxlnx cosx/x)/(1/y) dy/dx = xLxy ″ (x) y(x) ( s) = Lxcos2(x) ( s) = 1 2 ⋅ (Lx1 ( s) Lxcos(2x) ( s)) So, we use that Lxy ″ (x) ( s) = s2 ⋅ Y(s) − s ⋅ y(0) − y ′ (0) Lxy(x) ( s) = Y(s) Lx1 ( s) = 1 s Lxcos
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Then find a particular solution of the given nonhomogeneous equation x2y′′ −3xy′ 4y = x2 lnx, x > 0, y 1(x) = x2, y2(x) = x2 lnx Solution Notice that y′ 1(x) = 2x, and y′′ 1(xGraph y=x^2cos(x) Use the form to find the variables used to find the amplitude, period, phase shift, and vertical shift Find the amplitude Amplitude Find the period using the formula Tap for more steps The period of the function can be calculated using PeriodAnswer to y'' 4y' 4y = e^x cos x Find solutions for your homework or get textbooks Search
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优质解答 微分方程y″y=xcosx对应的齐次微分方程为y''y=0 特征方程为t 2 1=0 解得t 1 =i，t 2 =i 故齐次微分方程对应的通解y=C 1 cosxC 2 sinx 因此，微分方程y″y=xcosx对应的非齐次微分方程的特解可设为y * =axbx（csinxdcosx） y * '=acsinxdcosxcxcosxdxsinx y * ''=ccosxdsinxccosxcxsinxdsinxdxcosx 将y * ，y * '，y * ''代1 day ago · Show all steps find dy/dx by implicit differentiation 1 x sin y y sin x = 1 2 use inplicit diff find the equation of the tangent line to the curve at the given point sin(xy)=2x2y (pi,pi) 3 find y' and y'' y=ln(1lnx) 4 use log diff to findFind dy/dx x=cos(y) Differentiate both sides of the equation Differentiate using the Power Rule which states that is where Differentiate the right side of the equation Tap for more steps Differentiate using the chain rule, which states that is where and Tap for more steps
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Aug 02, 11 · Let x = y = 0 Then cos x = cos y = 1 But cos (0 0) does not equal cos 0 cos 0Solution f x x x cos y ye x cosy yex f y y x cosy ye x x sin y e x 2 f x 2 x f from MATH 2 at University of WollongongMar 11, 21 · Ex 55, 14 Find 𝑑𝑦/𝑑𝑥 of the functions in, 〖(cos〖𝑥 〗)〗^𝑦 = 〖(cos〖𝑦 〗)〗^𝑥Given 〖(cos𝑥)〗^𝑦 = 〖(cos𝑦
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First we find solution to the corresponding homogeneous equation math\qquad y'' 2y' y = 0/math by setting up and finding roots of characteristicMay 31, 10 · First find a solution to the homogeneous DE y'' y = 0 y=e^(rx) r² 1 = 0 r = / i y_h = C_1*sin(x) C_2*cos(x) Seek a particular solution to the nonhomogeneous DE
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