画像 (1-x^2)y'' 2xy'=0 y1=1 175216-(1-x^2)y''-2xy'+12y=0

Simplify x=y1 x=y1 Add y to both sides of the equation y^ {2}\left (2x\right)yx^ {2}1=0 All equations of the form ax^ {2}bxc=0 can be solved using the quadratic formula \frac {b±\sqrt {b^ {2}4ac}} {2a} The quadratic formula gives two solutions,Solution for ( 1 x2 ) y'' 2xy' 2y = 0 ;Find the specific solution for the following differential equations a) (dy/dx)^2 x^4 = 0 given the initial value y(1) = 0 b) dy/dx = ysinx given the initial value y(pi) = 1 calculus Sorry to post this again, but I am still unable to understand it and need help Please help1) Using 3(x3)(x^26x23)^2 as the answer to differentiating f(x

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(1-x^2)y''-2xy'+12y=0-Y1 = x Physics Social ScienceBy this expression, we can divide over the excess two X Y Q minus four Excuse over minus tree Extra square, wire square If we divide numerator and denominator by X, then one X will be canceled on In fact, we can also take minus sign out So we're left with four X square minus two Y cube over three X y square

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G@ Bð% Áÿ ÿ ü€ H FFmpeg Service01wThis problem has been solved!Divide \frac{y1}{y1}, the coefficient of the x term, by 2 to get \frac{y1}{2\left(y1\right)} Then add the square of \frac{y1}{2\left(y1\right)} to both sides of the equation This step makes the left hand side of the equation a perfect square

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(1−x2)y′′ −2xy′ 2y = 0, given that y1 = x is a solution SECTION 2 HIGHERORDER ODE'S 3 2C Secondorder Linear ODE's with Constant Coefficients 2C1 Find the general solution, or the solution satisfying the given initial conditions, to each of the following 1 Answers #1 1 y (1x^2)y'x (1y^2) = 0 this is a separable equation y ( 1 x^2)y' = x ( 1 y^2) rearrange as y / 1 y^2 y' = x / 1 x^2 integrate both sides ∫ y / 1 y^2 dy = ∫ x / 1 x^2 dx (1/2) ln ( 1 y^2) = (1/2) ln ( 1 x^2) CS '¼ õ†'rÙ² >TÀù ø 𠈩ºä~í@ŽbsŒ \âœÀ* O ŽÔ U &Ø"ì³Èå×nMÔ ÂX K ¨ÿ*„kÌŸéA ð¯vlvæÇd$0€y¬£õ€ô öÀ(H_Þr‡ÛFÀ= 51TDqœ‚oå4C6)> ¶§` Áó ©_ƺâ8œÇTž™Ç¢ÿÒÄŠíräãs ¾à¸b ž C·¢ y;

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 ⇒ (1 x 2)y 2 2xy 1 = 0 If y = ax b/ x^2 c, then (2xy1 y )y3 = A 3(xy2 y1)y2 B 3(xy2 y2)y2 C 3(xy2 y1)y1 D none of these asked in Derivatives by Ichha (27k points) derivatives;Answer (1 of 5) The equation \displaystyle{ (1x^2)y'' 2xy' 2y = 0 }\qquad(1) Since we have no obvious way to find any particular solution of (1) so we should try to find its general solution in the form of a power series as follows \displaystyle{ y = C_0 C_1x C_2x^2 \dots C_nx^2Dy dx P (x)y = Q(x) We have y' − 2xy = 1 with y(0) = y0 1 This is a First Order Ordinary Differential Equation in Standard Form So we compute and integrating factor, I, using;

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 Expert's answer T Solution (1x 2 )y'' 2xy' α (α 1)y = 0 Dividing both sides of the equation by (1x 2) we get differential equation in standard form y ′ ′ − 2 x 1 − x 2 y ′ α ( α 1) 1 − x 2 y = 0 y''\frac {2x} {1x^2}y'\frac {\alpha \left (\alpha 1\right)} {1x^2}y=0 y′′ − 1−Experts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high 92% (13 ratings)Solution 1 First assume y 2 =ux 2, and y 2 '=2uxu'x 2 and y 2 ''=u''x 2 4u'x2u Substituting into x 2 y''2xy'6y=0 we get x 2 (u''x 2 4u'x2u)2x(2uxu'x 2)6y=0 and simplifying by adding like terms we get u''x 4 6u'x 3 =0 We reduce the order by w=u' to get w'x 4 6wx 3 =0 Now dividing by wx 4 and rearranging, we get and integrating both

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Where m is the slope and b is the y intercept Divide each side by In other words, the tangent line is the graph of a locally linear approximation of the function near the point of tangency With this information, the tangent line has equation y=f'(c)(xc)f(c) Example 2 Find the equation of the normal line to the graph of at the point (−1, 2 We can use an integrating factor when we have a First Order Linear nonhomogeneous Ordinary Differential Equation of the form;`OHDR ?" ColumnsK ??

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X平方加y平方加2xy=c c为常数的图像时什么样的x2y22xy=c为什么呢 答案 两条互相平行,斜率为1的直线 补充回答原式可化简为(xy)^2=c xy=±√c y=x±√c 这是两个平行的直线方程 斜率都为1 在y轴的截距相错±√cPK !XáToa«, mimetypeapplication/epubzipPK !XáTò2©¯û METAINF/containerxmlMα  à½OAX LE7CJ›˜¸»øH¯•HïH £o/íÐtü/ÿŸïšî;yö 9BÎÇ Write an equation of a line that is parallel to the x axis and contains the point (7, 1 3) 1 puzzle Slopeintercept and pointslope form (3 1 Where does the line cross the yaxis and 2 what is the slope of the line Rlcraft Best Bow Enchants The slope of the line parallel to the given line is A line with a slope of 3 4 2 contains the point (8

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 find a series solution about the point x=0 of (1x^2)y"2xy'2y=0 find a series solution about the point x=0 of (1x^2)y"2xy'2y=0 1 Answer Sorted by 1 Alpha identifies it as Legendre's equation and gives the solution y ( x) = c 1 x c 2 ( − x ( log ( 1 − x) / 2 − log ( x 1)) − 1) It offers step by step if you have the right account ShareL ei6 H DP M Z 9 5 ݍ s CƂ ه m M yQ 68 0 I , y he xS, u ˺ b q1 m P P j u 7E ,d E v LF 9 m˦ ş KL ZU tV L ca wd HmQ Bܲ r

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=0 when =1 (1 x2) 2xy = 1 1 2 Divide both sides by (1 2) 2 1 2 = 1 1La nueva edición de esta prestigiosa obra conserva y refuerza la orientación de las aplicaciones a la administración y la economía, sin descuidar aplicaciones generales a otras áreas, como ciencias sociales, biológicas y físicas, con el propósito deI = e∫P (x)dx = exp(∫ −2x dx)

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@" b @= b ( r軥= O ( _H@= ?Can you solve this Q with steps please Show transcribed image text Consider the differential equation x^2y'' 2xy' 2y = 0, x > 0 (a) Show that y1 = x is a solution to the equation (b) Find another solution y2 such that y1 and y2 form a Details Calculator Use 3 Homogeneous Equations with Constant Coefficients y'' a y' b y = 0 where a and b are real constants $$ \begin {aligned} 3x 2y = & 1 \ 4x ~5y = & 14 \end {aligned} $$ Solution Step1 Multiply first equation by 5 and second by 2 Louis Arbogast introduced the differential operator (Abramowitz and Stegun 1972, p

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Solution for (1 – x)y" 2xy' = 0 , y1 close Start your trial now!H­´LÖôæ>ðéç èxžáÑŒl'Ä^ü}¹0M 2¼¡Üí_ óýU‚ã ÍÌ ŽÖò ‚ úKÄÿÁ~ Êâ,ÔA!5¨}g ‹ð ¾º ÈõÞ•¤£ õ¥/œu£Ù»Þ\r Ð )j k¶R¥'ö¸Ý\x¡€PRÜ£asH±¢ ¡¹{?Óî?ÛÜÎ%ÿº »!p´‡uÞÂ5"ýìø¤¢ÿÑJòðÁx ÁõùëO Scribd é o maior site social de leitura e publicação do mundo

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Incoming Term: (1-x^2)y''+2xy'=0 y1=1, (1-x^2)y''-2xy'+12y=0, solve 0=y=(1-x^(2))y''-2xy'+1(2+1)y,