Guide :

# solve the differential equation y''=x^-4

solve by the simple deferntional method

## Research, Knowledge and Information :

### Second Order Linear Differential Equations - University of Utah

y y . We have already seen (in section 6.4) how to solve ﬁrst order linear equations; ... x which solves the differential equation (12.1) ...
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### solve differential equation $y' + (\cos x)y = 4 \cos x ... ... y = 4 \cos x$$This can be done using the differential ... solve differential equation y' + (\cos x)y = 4 \cos x ... If I could get a nice way to solve$$ \int ... Read More At : math.stackexchange.com... ### How to Solve Differential Equations A DIFFERENTIAL EQUATION ... Many differential equations may be solved by separating the variables x and y on opposite sides of the ... Solve : xy' = y 2: Solution : ... Read More At : math.uww.edu... ### 1. Solving Differential Equations - IntMath 1. Solving Differential Equations ... It is the same concept when solving differential equations ... After solving the differential equation, (dy)/(dx)ln\ x-y/x=0 Read More At : www.intmath.com... ### Solve The Given Differential Equation. Y' = X^4/y | Chegg.com Answer to Solve the given differential equation. y' = x^4/y... Read More At : www.chegg.com... ### Solving the differential equation$y'= 2 xy + 4 x$using ... Solving the differential equation$y'= 2 xy + 4 x\$ using power series. up vote 2 down vote favorite. 1. ... Solve a differential equation using the power series ...
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### How to Solve Differential Equations - wikiHow

How to Solve Differential Equations. ... We do not solve partial differential equations in this article because the methods for solving these types of equations ...
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### First-Order Linear Di erential Equations

First-Order Linear Di erential Equations: AFirst order linear di erential equationis an equation of the form y0+P(x) ... of y. We can solve this equation in general ...
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### Separable equations example (old) (video) | Khan Academy

An old worked example video of solving separable equations. ... There isn't just one tool or one theory that will solve all differential equations.
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### Using Series to Solve Differential Equations

Using Series to Solve Differential Equations Many differential equations can’t be solved explicitly in terms of ﬁnite combinations of simple familiar functions.
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## Suggested Questions And Answer :

### What is the general equation of the differential equation d^2/dx^2 + 4y = 0 ?

I read this as y"+4y=0 where y" is d^2y/dx^2 and y' is dy/dx. We start with the simple fact that the differential of sine is cosine and the differential of cosine is minus sine, so we have a repeating pattern in the differential. Furthermore, when we differentiate twice we also get a change of sign from which we can see that the double differential of sine or cosine is minus sine or cosine. This is the main clue in solving the differential equation. Let y=bsin(ax), y'=abcos(ax) and y"=-a^2bsin(ax)=-a^2y. So y"+a^2y=0. (Notice that the constant b is carried through unaffected by the differentiation.) Comparing this with y"+4y=0, a^2=4 so a is -2 or +2.  Check: a=2 y=bsin(2x) y'=2bcos(2x) y"=-4bsin(2x)=-4y, y"+4y=0 a=-2 y=bsin(-2x)=-bsin(2x) y'=-2bcos(2x) y"=4bsin(2x)=-4y, y"+4y=0 However, the same result applies if y=bcos(2x). If we have two functions y1 and y2 we can put y=y1+y2. It's true that y'=y1'+y2' and y"=y1"+y2". So we can let y1=bsin(2x) and y2=ccos(2x), making y=bsin(2x)+ccos(2x); y"=-4bsin(2x)-4ccos(2x)=-4y, so the general answer is y=Asin(2x)+Bcos(2x), where A and B are constants.
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### Solve the equation p^2+p(x+y)+xy=0

Question:  Solve the equation p^2+p(x+y)+xy=0. Its related to engineering math -  Clairaut's equation. Since Clairaut's equation is a differential equation, then I am assuming that p means the first differential, dy/dx, or y'. You equation actually then is: (y')^2 + (y')*(x+y) + xy = 0. Treating your equation as a simple quadratic equation, and using the quadratic formula to solve it, p = {-(x+y) +/- sqrt((x+y)^2 - 4xy)}/(2*1) p = {-(x+y) +/- sqrt((x-y)^2)}/2 p = {-(x+y) +/- (x-y)}/2 y' = {-(x+y) - (x-y)}/2,  y' = {-(x+y) + (x-y)}/2, y' = -x,   y' = -y And from these the solutions are, y1(x) = C - (1/2)x^2,  y2(x) = ke^(-x)
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### what is general solution for dw/dt = 7w+6y-10z

what is general solution for dw/dt = 7w+6y-10z dx/dt= 3w + x +3y-5z dy/dt= 5w+2x+11y-14z dz/dt=5w+2x+8y-11z Represent the coupled differential equations in matrix form. X = Av Where X is the column vector [w’ x’ y’ z’] and x’ = dx/dt, etc. And A is the component matrix, |7  0   6  -10| |3  1   3    -5| |5  2 11  -14| |5  2   8   -11| And v is the column vector [w x y z] To find the solutions to the original coupled differential equations we need to solve Av = λv for eigenvalues and eigenvectors. i.e. det(A –λI) = 0 or, |7-λ  0     6      -10| = 0 |  3  1-λ   3        -5| |  5    2  11-λ   -14| |  5    2    8    -11-λ| (7-λ){(1-λ)[(11-λ)(-11-λ) – (-14)(8)] – 3[2(-11-λ) – (-14)(2)] + (-5)[2*8 – (11-λ)*2]} – 0 + 6{3[2(-11-λ) – (-14)*2] – (1-λ)[5*(-11-λ) – (-14)*5] + (-5)[5*2 – 2*5]} – (-10){3[2*8 – (11-λ)2] – (1-λ)[5*8 – (11-λ)*5] + 3[5*2 – 2*5]} = 0 You could evaluate and solve the above expression for λ manually, or use an algebra software package such as Maple, Matlab or Mathematica. The solutions for λ are, λ = 1, λ = 1, λ = 3, λ = 3 i.e. four solutions with two roots repeated twice. The eigenvectors are got by solving Av = λv, for λ = 1, 3  i.e. |7  0   6  -10 ||w| = |w|   and   |7  0   6  -10 ||w| = |3w|   |3  1   3    -5 || x|     |x|              |3  1   3    -5 || x|     |3x| |5  2 11  -14|| y|     | y|             |5  2 11  -14|| y|     | 3y| |5  2   8   -11|| z|     | z|             |5  2   8   -11|| z|     | 3z| Giving, 7w + 6y – 10z = w                 and       7w + 6y – 10z = 3w 3w + x + 3y – 5z = x                            3w + x + 3y – 5z = 3x 5w + 2x + 11y – 14z = y                      5w + 2x + 11y – 14z = 3y 5w + 2x + 8y – 11z = z                         5w + 2x + 8y – 11z = 3z   The solution of which is:                    The solution of which is: w = 2, x = 1, y = 3, z = 3                       w = 2, x = 1, y = 2, z = 2 i.e. v1 = [2, 1, 3, 3]                               i.e. v2 = [2, 1, 2, 2] Independent solutions of the coupled equations then are, x1 = v1.e^t,   x3 = v2.e^(3t)   Repeated roots λ = 1, and one solution is x1. A 2nd solution is x2 = t.x1 + p.e^t Where p is solved from (A – λI)p = v1 | 6  0   6   -10||p1| = |2| | 3  0   3     -5||p2|    |1| | 5  2  10 -14||p3|    |3| | 5  2   8  -12||p4|    |3| Giving, p = [p1, p2, p3, p4] = [k, k, k, k] Taking k = 1, p = [1, 1, 1, 1] Then x2 = t.v1.e^t + p.e^t   λ = 3, and one solution is x3. A 2nd solution is x4 = t.x3 + q.e^(3t) Where q is solved from (A – λI)q = v2 | 4   0   6   -10||q1| = |2| | 3  -2   3     -5||q2|    |1| | 5   2   8   -14||q3|    |2| | 5   2   8   -14||q4|    |2| Giving, q = [q1, q2, q3, q4] = [2k, k, 2k+ 2, 2k + 1] Taking k = 0, q = [0, 0, 2, 1] Then x4 = t.v2.e^(3t) + q.e^(3t) Our general solution then is: x = c1.x1 + c2.x2 + c3.x3 + c4.x4 x = c1.v1.e^t + c2{ t.v1.e^t + p.e^t } + c3.v2.e^(3t) + c4{ t.v2.e^(3t) + q.e^(3t) }
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### How do I solve equations simultaneously?

In mathematics, simultaneous equations and systems of equations are finite sets of equations whose common solutions are looked for. The systems of equations are usually classified in the same way as the single equations, namely: System of linear equations System of polynomial equations System of ordinary differential equations System of partial differential equations
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### Solve the initial value problem dy/dx + 4xy = e−2x with y(0) = 1

Let Y,h: Y'h+4xY,h=0
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### If the equation is non homogeneous differential equation, is the solution exist?

The simple answer is that it often does, because the inhomogeneous DE can be split into a homogeneous part solvable by conventional means (e.g., solving the characteristic equation), and a particular solution by examining what remains when the homogeneous part is removed (and solved).
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### (1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

### DIFFERENTIAL EQUATIONS

Verify eg range’s identity for the equation   x2y1+7xy’+8y=0 ?Verify eg range’s identity for the equation   x2y1+7xy’+8y=0 ?
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### solve differential equation: y' + y^2 sinx = 0

Question: solve differential equation: y' + y^2 sinx = 0 dy/dx = -y^2.sin(x) int dy/y^2 = -int sin(x) dx -1/y = C + cos(x) y(x) = -1/(C+cos(x))
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