[最も好ましい] y' 4xy=x^3e^x^2 y(0)=-1 274644

Since y = v1/3, y = 3 r C x2 3x (f) y0ycot x = cos x Solution We see that this equation is already linear, with p(x) = cot x = cos x sin x, so its integrating factor is m(x) = e R cot xdx = elnjsin xj = sin x Multiplying by this integrating factor, we get (ysin x)0= sin xcos x, so integrating gives ysin x = 1 2 sin 2 x C, and y = 1 2 sin xX dy/dx y = 4x 1 y(1) = 8 y' 4xy = x^3 e^x^2 y(0) = 1 (x1) dy/dx y ln x y(1) = 10 x(x 1) dy/dx xy = 1 y(e) = 1 y' (sin x)y = 2 sin x y(/2) = 1 y' (tan x)y = cos^2 x y(0) = 1 In Problems 3740 proceed as in Example 6 to solve the given initialvalue problem Use a graphing utility to graph the continuous function y(x)1/2yx5/2=0;y2x3=0 No solution System of Linear Equations entered 1 1/2yx5/2=0 2 y2x3=0 Equations Simplified or Rearranged 1 y/2 x = 5/2 2 y 2x =

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Y' 4xy=x^3e^x^2 y(0)=-1-\bold{\mathrm{Basic}} \bold{\alpha\beta\gamma} \bold{\mathrm{AB\Gamma}} \bold{\sin\cos} \bold{\ge\div\rightarrow} \bold{\overline{x}\space\mathbb{C}\forall}Extended Keyboard Examples Upload Random 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

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A) x2 y′′ 4xy′ 2 y =0 b) x2 y′′ 5 xy′ 4 y =0 c) 2 x2 y′′ 3xy′ 4 y =0 Solution a) Set y= xr we reach r (r − 1)4r 2=0 r1, 2 = −2, −1 (1) So the general solution is y= C1x− 2 C2 x− 1 (2) b) Set y= xr we get the indicial equation r (r − 1)5 r 4=0 r1, 2STAT 400 Joint Probability Distributions Fall 17 1 Let X and Y have the joint pdf f X, Y (x, y) = C x 2 y 3, 0 < x < 1, 0 < y < x, zero elsewhere a) What must the value of C be so that f X, Y (x, y) is a valid joint pdf?b) Find P (X Y < 1)c) Let 0 < a < 1 Find P (Y < a X) d) Let a > 1 Find P (Y < a X)e) Let 0 < a < 1 Find P (X Y < a)Answered 9 months ago Author has 12K answers and 3529K answer views I think you forgot the sign =0 If so, write the equation as M (x,y)dx N (x,y)dy =0 , with M = y (x^3e^xy y) , M_y = x^3e^xy yx^4e^xy 2y N = x (y x^3e^xy) , N_x = 4x^3e^xy yx^4e^xy y # M_y

1115 Substitute y = erx into the given differential equation to determine all values of the constant r for which y = erx is a solution of the equation y′′ y′ − 2y = 0 Solution The first and second derivatives of y = erx are y′ = rerx, y′′ = r2erx If we plug these into the differential equation we get r2e rxrerx − 2e = (r2 r −2)erx = 0 As erx 6= 0 for any r or xREDUCCION DE ORDEN E100 5´ Si y2(x)=u(x)y1(x) ⇒ y2 = ue−2x ⇒ y 0 2 = u e −2x − 2ue−2x ⇒ y00 2 = u 00e−2x − 4u0e−2x 4ue−2x Sustituyendo en (2x 1)y00 2 4xy 0 2 − 4y2 =0 Se obtiene (2x1)(u00 − 4u0 4u)e−2x 4x(u0 − 2u)e−2x − 4ue−2x =0 Multiplicando por e2x se tiene que (2x 1)(u00 − 4u0 4u)4x(u0 −2u)−4u =0 (2x 1)u00 (−8x − 44x)u0 (8x4−8xSee the answer See the answer done loading In Exercises 7–29 use variation of parameters to find a particular solution, given the solutions y1, y2 of the complementary equation x^2y'' − 4xy' 6y = x^5/2 , x > 0;

There is a standard procedure to convert a linear differential equation of order > 1 into a system of first order linear differential equations and the number of equations and of the unknown functions is same as the order of the given equation y'' (4x/x^2 1)y' (2/x^2 1)y =0 Now put y1) (sec2 xy4 e3y sinx) 4xy3 3e3y cosx 9y2 1y3 dy dx = 0 M (x;y)dxN (x;y)dy = 0 @M(x;y) @y = @(sec 2 xy4 3e y sinx) @y = 04y3 3e3y sinx;Step 1 1 of 5 (a) We can detect the linear dependency of the functions by graphing them \\\\ all on the coordinate axes, and see if they are multiple of each other or not \\\\ Below are the graphs of both the functions y 1 = x 3 \\\\ {\color {#c} y_1=x^3}\quad y 1 = x 3 and

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18 Resolva os problemas de valor inicial a seguir (a) y′ y =xex, y(0)=0 Resp fator integrante ex, solução geral y =x−1ex 2 Ce −x, solução y =x−1exe−x 2 (b) xy′ 2y =x3, y(1)=0, x > 0 Resp fator integrante x2, solução geral y = x3 5 C x2, solução y = x3 5 − 1 5x2 (c) y′ −2xy =3x2ex2, y(0)=1 Resp fator integrante e−x2, solução geral y =(x3 C)ex2 Section 43 Double Integrals over General Regions In the previous section we looked at double integrals over rectangular regions The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral, ∬ D f (x,y) dA ∬ D f ( x, y) d A where D D is any regionTHEOREM 2 If y = y1(x) and y = y2(x) are any two solutions of (H), then u(x)=y1(x)y2(x) is also a solution of (H) Proof Let y= y1(x) and y= y2(x) be any two solutions of (H)Then y00 1(x)p(x)y0 1(x)q(x)y1(x) = 0 and y00 2(x)p(x)y0 2(x)q(x)y2(x)=0 Now set u(x)=y1(x)y2(x)Then u(x)=y1(x)y2(x) u0(x)=y0 1(x)y02 (x) u00(x)=y00 1(x)y00 2(x) Substituting u into (H), we get

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2 3 (e−1)(8−1) = 14 3 (e−1) 4 09 Evaluating the limits of integration When evaluating double integrals it is very common not to be told the limits of integration but simply told that the integral is to be taken over a certain specified 0 4xy 2yy=2x y=x2 dx = Z 2 0 8x2 4x y = (x43)e^(2x^2) We have y' 4xy = e^(2x^2) A We can use an integrating factor when we have a First Order Linear nonhomogeneous Ordinary Differential Equation of the form;5 (e y1)2 e−dx (e x1)3 e− dy =0 6 ¡ y −yx2 ¢ dy dx =(y 1)2 7 dy dx =sinx ¡ cos2y −cos2 y ¢ 8 x p 1−y2 dx = dy 9 (e xe−) dy dx = y2 (21) ds dr = ks, 1 s ds = kdr, Z 1 s ds = k Z dr, lns = krc 1, s = ekrc 1 s = ekrec 1 = c 2ekr, (c 2 = ec 1), s = ±c 2ekr, s

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If two random variable has joint pdf f(x, y) = (6/5) (x y 2), 0 < x < 1 , 0< y5 Substitute u back into the equation we got at step 2;Answer to Solve the initialvalue problem y' 4xy = x^3 e^{x^2}, y(0) = 8 By signing up, you'll get thousands of stepbystep solutions to for Teachers for Schools for Working Scholars® for

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Y2 2 y3 3 − 7y4 12 1 0 = 2 3 (e) La regi´on de integracion es la que se ilustra en la figura adjunta 2 8 y=4x y=x3 Intercambiando el orden de integracion se obtiene I = Z 2 0 dx Z 4x x3 ex2dy = Z 2 0 (4xex2 −x3ex2)dx = 2ex2 2 2 0 − x2 2 ex2 0 Z 0 xex2dx = e4 2 − 5 2 (Aplicar el m´etodo de integraci´on por partes en la segundaDy/dx P(x)y=Q(x) The given equation is already in standard form, so the integrating factor is given by;3 Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step) 4 Solve using separation of variables to find u;

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FirstOrder Linear Equations Solve firstorder linear equations y' (t) 2y (t) = 3 e^ (2t) x y' (x) 4 y (x) = x^6 exp (x), y (1) = 0 See steps that use Laplace transforms to solve an ODE solve y' (t) 3y (t) = delta (t 2), where y (0) = 0 More examples RELATED EXAMPLES Differential EquationsFactor out the Greatest Common Factor (GCF), 'd' d(6xX 4xy 3 X 3x 2 yX 6x 2 y 3 x 3 y 1y) = 0 Subproblem 1 Set the factor 'd' equal to zero and attempt to solve Simplifying d = 0 Solving d = 0 Move all terms containing d to the left, all other terms to the rightSection 31 SecondOrderLinear Equations 311 Verify that the functions y 1 and y 2 given below are solutions to the secondorder ODE also given below Then, find a particular solution of the form y = c 1y 1 c 2y 2 that satisfies the given initial conditions Primes denote derivatives with respect to

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( x;y) = R N (x;yY1 = x^2 , y2 = x^3 4xy'' 2y' y = sin sqrt (x);1) is a critical point The second derivative test f xx = 2;f yy = 2;f xy = 0 shows this a local minimum with

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1 Substitute y = uv, and dy dx = u dv dx v du dx into dy dx P(x)y = Q(x) 2 Factor the parts involving v;F(02,03) 6= fX(02)fY (03) (e) E(eX cosY) = Z 1 0 Z 2 0 (ex cosy)(x2 xy 3)dydx (f) cov(X,Y) = E(XY)−E(X)E(Y) = Z 1 0 Z 2 0 xy(x2 xy 3)dydx− Z 1 0 Z 2 0 x(x2 xy 3)dydx Z 1 0 Z 2 0 y(x2 xy 3)dydx Example 2 X and Y are jointly continuous with joint pdf f(x,y) = ˆ cxy if 0 ≤ x, 0 ≤ y, xy ≤ 1 0, otherwise (a) Find c (bIf a function is written in the form \frac{dy}{dx}=f(x)g(y), rearrange the function and write it as \frac{1}{g(y)}dy=f(x)dx Then both sides can be integrated to solve for y The definite integral can be solved using that equation

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0 1 y = x 1 3 e3x C 0 4 Solve for y 1 y = x 1 3 e3x C y = 1 1x 3 e3x C 5 Apply the initial condition y(0) = 1 and solve for C 1 = 1 10 3 e3(0) C 1 1 = 1 3 e3(0) C 1 = 1 3 C 2 3 = C 6 Check for constant singular solutions 0 = y2 e3xy2 = y2(1 e3x) So y = 0 is a solution to the DE However, it doesn't satisfy the initialFind dy/dx x^24xyy^2=4 Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is whereЛинейное однородное уравнение какого порядка на интервале (0, 1) может иметь такие четыре частных решения y 1 = x 2 2x 2, y 2 = (x 2) 2,

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Y1 = e^x , y2 = x^3e^x@N(x;y) @x = @ 4xy 33e y cosx 9y 2 1y3 @x = 4y3 3e3y sinx0) @M(x;y) @y = @N(x;y) @x ) Exact DE ( x;y) = R M (x;y)dx = R (sec2 xy4 e3y sinx)dxA(y) = tanxxy4 e3y cosxA(y);Y1 = cos sqrt (x), y2 = sin sqrt (x) xy'' − (2x 2)y' (x 2)y = 6x^3e^x ;

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Proposition 1 For A ⊂ R2, P((X,Y) ∈ A) = Z Z A f(x,y)dxdy The twodimensional integral is over the subset A of R2Typically, when we want to actually compute this integral we have to write itSolution for X^24xyy^2=0 equation Simplifying X 2 4xy y 2 = 0 Solving X 2 4xy y 2 = 0 Solving for variable 'X' Move all terms containing X to the left, all other terms to the right Add '4xy' to each side of the equationLa solución queda expresada con la siguiente regla de correspondencia 1 ;

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0 23k views Solve ( x y 2 − e 1 / x 3) d x − x 2 y d y = 0 written 54 years ago by aksh_31 ♦ 23k • modified 54 years ago Mumbai University > First Year Engineering > sem 2 > Applied Maths 2 Marks 6 Year 14 needtagging a) x 2xy nessa você vai apenas substituir os valores de acordo com cada incógnita na questão diz que x é equivalente a 3 e y a 1, então, onde tiver x você coloca o 3 e onde tiver y o 1 então vamos ter3 2(3)1 lembrando de manter o sinal quando for substituirI = e^(int P(x) dx) \ \ = exp(int \ 4x \ dx) \ \ = exp( 2x^2 ) \ \ = e^(2x^2) And if we multiply

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 Q565 36 Suppose p 1 and p 2 are continuous on ( a, b) Let y 1 be a solution of (A) y ″ p 1 ( x) y ′ p 2 ( x) y = 0 that has no zeros on ( a, b), and let x 0 be in ( a, b) Use reduction of order to show that y 1 and (56E1) y 2 ( x) = y 1 ( x) ∫ x 0 x 1 y 1 2 ( t) exp ⁡Example 7 y = C1x2 C2x2x3 is the general solution of the second order differential equation x 2 y 00 −2xy 0 2y =4x 3 Setting C 1 = 2 and C 2 = 3, we 32 y'4xy=x^3 e^{x^2}, y(0)=1 Ecuaciones lineales Alexander Estrada

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1 2 x2ex 11 y00 y0 1 4 y = 3 e 1 2 x Sol The characteristic equation m2 m 1 4 = (m 1 2) 2 = 0 has a root m = 1 2 with multiplicity 2 The complementary solution is y c = C 1e 1 2 x C 2xe 1 2 x In view of Superposition Principle, we seek a particular solution y p = y p 1 y p 2 where y p 1 and y p 2 are particular solutions of y00 y032 y'4xy=x^3 e^{x^2}, y(0)=1 Ej 23 Parte 2 Explicación de la integral Ecuaciones linealesLink de descarga del librohttps//1drvms/b/s!AtFwQmgnspIqCalculadora gratuita de ecuaciones diferenciales ordinarias (EDO) Resolver ecuaciones diferenciales ordinarias paso por paso

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Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange6 Solve that to find v;Subject to the constraint 2x2 (y 1)2 18 Solution We check for the critical points in the interior f x = 2x;f y = 2(y1) =)(0;

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I'm working on finding the general solution of $(x^21)y''4xy'2y=0$ in powers I assume the form $$ y(x)=\sum_{n=0}^\infty C_nx^n$$ My basic strategy is to first figure out each piece individually, shift the indexes if necessary, and then rewrite it into one seriesIn fact, as a particular solution of y′′ 2y′ y= e−t 9t we can take 1 9ln(t)te−9t because the term −1 9te −t is included in the solution space of the corresponding homogeneous equation It follows that the general solution of the differential equation y′′ 2y′ y= e−t 9t (t > 0) is y(t) = C 1e−t C 2te−t 1 9 ln(t)te−t where C9 Find general solution of the following di erential equations given a known solution y 1 (i) (T) x(1 0x)y00 2(1 2x)y 2y= 0 y 1 = 1=x (ii) (1 2x)y00 2xy0 2y= 0 y 1 = x Solution (i) Here y 1 = 1=x Substitute y = u(x)=xto get (1 x)u00 2u0= 0 Thus, u0= 1=(1 x)2 and u= 1=(1 x) Hence, y 2 = 1=(x(1 x)) and the general solution is y= a=x b=(x

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