[go: up one dir, main page]

MaplePrimes Questions

Search Questions:

how this integro-differential equation can be solved?
any assumption or suggestion is appreciated. tnx in advance

restart

eq:=diff(y(x),x)=y(x)+exp(x)*exp(-3*x)/2+int(exp(x+t)*y(t),t=0..x);

IC:=y(0)=1;

diff(y(x), x) = y(x)+(1/2)*exp(x)*exp(-3*x)+int(exp(x+t)*y(t), t = 0 .. x)

 

y(0) = 1

(1)

eq2:=g(x)=int(exp(x+t)*y(t),t=0..x);

g(x) = int(exp(x+t)*y(t), t = 0 .. x)

(2)

IC_2:=eval(eq2,[x=0,y=1])

g(0) = 0

(3)

sys:={diff(eq2,x),subs(rhs(eq2)=lhs(eq2),eq)}

{diff(g(x), x) = int(exp(x+t)*y(t), t = 0 .. x)+exp(2*x)*y(x), diff(y(x), x) = y(x)+(1/2)*exp(x)*exp(-3*x)+g(x)}

(4)

dsolve(sys union {IC,IC_2},numeric)

Error, (in dsolve/numeric/process_input) input system must be an ODE system, got independent variables {t, x}

 

Download integro-diffrential_problem.mw

Hey all Maple experts.I could really use some help with  diff,D,Diff

restart

interface(version)

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

(1)

with(Physics[Vectors])

NULL

CompactDisplay(A_(x, y, z, t), `ϕ`(x, y, z, t), v_(x, y, z, t), F_(x, y, z, t), v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t), A__x(x, y, z, t), A__y(x, y, z, t), A__z(x, y, z, t), quiet)

macro(Av = A_(x, y, z, t), `ϑ` = `ϕ`(x, y, z, t), Vv = v_(x, y, z, t), Fv = F_(x, y, z, t))

show, ON, OFF, kd_, ep_, Av, vartheta, Vv, Fv

(2)

Fv = q*('-VectorCalculus[Nabla](`ϑ`)'-(diff(Av, t))+`&x`(Vv, `&x`(VectorCalculus[Nabla], Av)))

F_(x, y, z, t) = q*(-Physics:-Vectors:-Nabla(varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics:-Vectors:-`&x`(v_(x, y, z, t), Physics:-Vectors:-Curl(A_(x, y, z, t))))

(3)

Av = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, Vv = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

(4)

subs[eval](A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k, F_(x, y, z, t) = q*(-Physics[Vectors][Nabla](varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics[Vectors][`&x`](v_(x, y, z, t), Physics[Vectors][Curl](A_(x, y, z, t)))))

F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k)

(5)

map(Component, F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k), 1)

F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q

(6)

collect(F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q, [q, v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t)])

F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q

(7)

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, Diff)

F__x = (v__y(x, y, z, t)*(Diff(A__y(x, y, z, t), x)-(Diff(A__x(x, y, z, t), y)))+(Diff(A__z(x, y, z, t), x)-(Diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(Diff(varphi(x, y, z, t), x))-(Diff(A__x(x, y, z, t), t)))*q

(8)

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, D)

F__x = (v__y(x, y, z, t)*((D[1](A__y))(x, y, z, t)-(D[2](A__x))(x, y, z, t))+((D[1](A__z))(x, y, z, t)-(D[3](A__x))(x, y, z, t))*v__z(x, y, z, t)-(D[1](varphi))(x, y, z, t)-(D[4](A__x))(x, y, z, t))*q

(9)

 
Hello everyone, in the result of this command execution process, it appears that the symbols for partial derivatives and derivatives in equation (8) are displayed incorrectly. What should I do?

Download error_display.mw

I am trying to compute partial derivatives of some complicated expression which include summations. First, I noticed that sum behaves differently if I use 1D vs. 2D math. Why?

Questions:
  1. Partial derivative of a summation: why is it not just 2*X[i]?
  2. Partial derivative of a double summation: how to define the nested structure of a double summation with j<>i?
  3. System of n+1 equations: how to define and solve for it?

For 3., each i equation is the partial derivative of my complicated expression with summations with respect to X[i], where i ranges from 1 to n. The last equation is the partial derivative with respect to X_r (a fixed variable).

Thanks.

restart

A := sum(X[i]^2, i = 1 .. n); eq[1] := diff(A, X[i]) = 0

sum(X[i]^2, i = 1 .. n)

 

sum(2*X[i], i = 1 .. n) = 0

(1)

B__wrong := sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n); B__correct := 'sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)'

n^2*X__r^2*w^2+sum(sum(X__r*w*X[j]+X[i]*X[j], j = 1 .. n)+n*X__r*w*X[i], i = 1 .. n)

 

sum(sum((X__r*w+X[i])*(X__r*w+X[j]), j = 1 .. n), i = 1 .. n)

(2)

eqs := seq(eq[i], i = 1 .. n); vars := seq(X[i], i = 1 .. n)

Error, range bounds in seq must be numeric or character

 

Error, range bounds in seq must be numeric or character

 
 

NULL

Download equations_with_summations.mw

On some occasions I have seen output like this

where the user name is replaced by a placeholder.
How to achieve this?

Let P(u,v) = -u⁴+88u³v-146u²v²+88uv³-v⁴+2u²+40uv+2v²-1, with P(u,v)=0,

and u0 = sqrt(2)/6 i,  v0 = -sqrt(2)/6 i. We see that P(u0,v0)=0.

Direct substitution of implicitdiff(P(u,v),v,u) at u0, v0 leads to 0/0.

For obtaining the true value of dv/du at (u0,v0) we have written in Maple:

solve(limit(subs(v=v0+k*(u-u0), implicitdiff(P(u,v),v,u)), u=u0)-k, k);

There are two solutions: 49/113 - 72*sqrt(2)/113*I, and 49/113 + 72*sqrt(2)/113*I.

However we were unable to determine the true value of d²v/du² at (u0,v0).

QUESTION:  How to determine it?

Hi. I'm comparing two 7x3-man teams lists. I can see there are at least 2 teams that have common members, new[1] and past[4], new[4] and past[3]. There may be others but the chat gpt code below doesn't definitively find any of them.

common.mw

The company library with all the good stuff has passed a 100 Mb size, and I wonder if it is possible to find out, which parts of the library that use most of the storage space?

After exertion with ordinary differential equations now relaxation:

Determine the formation law, limit and sum limit for
u_n+3=(13/12)*u_n+2 - (3/8)*u_n+1 + (1/24)*u_n .
Starting values ​​u_1=0, u_2=1, u_3=1.

I am trying to solve my first-order ODE by using BCs. I need explicit relation. Could anyone have a look at my maple work?

Help_explicitly_Solution.mw

If we calculating it take to much time but if we make a procedure it will be more effectable for such example, i want the exact and approximat and error

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(inttrans)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

eq := diff(y(x, t), `$`(t, 2))+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

diff(diff(y(x, t), t), t)+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

(2)

eq1 := laplace(eq, t, s)

s^2*laplace(y(x, t), t, s)-s*y(x, 0)+laplace(diff(y(x, t), x), t, s)*x-laplace(y(x, t), t, s)-(D[2](y))(x, 0)+laplace(diff(y(x, t), x), t, s) = 2*laplace(y(x, t)^3, t, s)

(3)

eq2 := subs({y(x, 0) = 1, (D[2](y))(x, 0) = 1}, eq1)

s^2*laplace(y(x, t), t, s)-s+laplace(diff(y(x, t), x), t, s)*x-laplace(y(x, t), t, s)-1+laplace(diff(y(x, t), x), t, s) = 2*laplace(y(x, t)^3, t, s)

(4)

eq3 := s^2*laplace(y(x, t), t, s) = s-laplace(diff(y(x, t), x), t, s)*x+1+laplace(diff(y(x, t), x), t, s)+2*laplace(y(x, t)^3, t, s)+laplace(y(x, t), t, s)

s^2*laplace(y(x, t), t, s) = s-laplace(diff(y(x, t), x), t, s)*x+1+laplace(diff(y(x, t), x), t, s)+2*laplace(y(x, t)^3, t, s)+laplace(y(x, t), t, s)

(5)

eq4 := expand(eq3/s^2)

laplace(y(x, t), t, s) = 1/s-laplace(diff(y(x, t), x), t, s)*x/s^2+1/s^2+laplace(diff(y(x, t), x), t, s)/s^2+2*laplace(y(x, t)^3, t, s)/s^2+laplace(y(x, t), t, s)/s^2

(6)

NULL

"u[0](x):=invlaplace(1/s+1/(s^2),s,x)"

proc (x) options operator, arrow, function_assign; invlaplace(1/s+1/s^2, s, x) end proc

(7)

u[0](x)

1+x

(8)

n := N

N

(9)

k := K

K

(10)

f := proc (u) options operator, arrow; u^3 end proc

proc (u) options operator, arrow; u^3 end proc

(11)

for j from 0 to 3 do A[j] := subs(lambda = 0, (diff(f(seq(sum(lambda^i*u[i](x), i = 0 .. 20), m = 1 .. 2)), [`$`(lambda, j)]))/factorial(j)) end do

(1+x)^3

 

3*(1+x)^2*u[1](x)

 

3*(1+x)*u[1](x)^2+3*(1+x)^2*u[2](x)

 

u[1](x)^3+6*(1+x)*u[1](x)*u[2](x)+3*(1+x)^2*u[3](x)

(12)

A[0]

(1+x)^3

(13)

y[0] := 1+x

1+x

(14)

y[1] := invlaplace(2*laplace(A[0], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[0], x), x, s)/s^2, s, x)+invlaplace(laplace(y[0], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[0], x), x, s)/s^2, s, x)

(1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2

(15)

y[1] := expand((1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2)

(1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2

(16)

"u[1](x) :=y[1]  "

proc (x) options operator, arrow, function_assign; y[1] end proc

(17)

NULL

A[1]

3*(1+x)^2*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)

(18)

y[2] := invlaplace(2*laplace(A[1], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[1], x), x, s)/s^2, s, x)+invlaplace(laplace(y[1], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[1], x), x, s)/s^2, s, x)

(1/840)*x^4*(7*x^5+63*x^4+212*x^3+476*x^2+672*x+420)-(1/60)*x^4*(x^3+6*x^2+10*x+20)+(1/420)*x^4*(x^3+7*x^2+14*x+35)-(1/60)*x^3*(x^3+6*x^2+10*x+20)

(19)

y[2] := expand(%)

(1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3

(20)

" u[2](x):=y[2]"

proc (x) options operator, arrow, function_assign; y[2] end proc

(21)

A[2]

3*(1+x)*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)^2+3*(1+x)^2*((1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3)

(22)

y[3] := invlaplace(2*laplace(A[2], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[2], x), x, s)/s^2, s, x)+invlaplace(laplace(y[2], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[2], x), x, s)/s^2, s, x)

(1/10810800)*x^5*(7623*x^8+99099*x^7+518778*x^6+1634490*x^5+3647930*x^4+5167305*x^3+4221360*x^2+900900*x-1081080)-(1/25200)*x^5*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)+(1/831600)*x^5*(63*x^6+693*x^5+2750*x^4+6930*x^3+11220*x^2+2310*x-13860)-(1/25200)*x^4*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)

(23)

y[3] := expand(y[3])

(286/945)*x^9+(131/336)*x^8+(1/7)*x^10+(17/70)*x^7-(1/40)*x^6-(1/20)*x^5+(1/12)*x^4+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

(24)

NULL

addingterm := y[0]+y[1]+y[2]+y[3]

1+x+(37/60)*x^5+(2/3)*x^4+(1/3)*x^3+x^2+(2351/7560)*x^9+(781/1680)*x^8+(101/210)*x^7+(53/120)*x^6+(1/7)*x^10+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

(25)


 

Download aproximate_and_exact_solution.mw

a table like that

 

there is four formula for calculate them which i know them by name of author the first one is adomian second one is (BiazarShafiofAdomian) which one member of mableprimes write code for me,but i don't know how use for all kind function maybe in future i upload this program for fix this issue, the third one is by zhao which is i think i easy for calculate just  i need someone one to wite the program and do some test for some example i  upload some picture in case for getting algorithm to writting and have some example for testing  so  lets see who can do this algorithm is very usfule when we solve ODE or PDE by LDM, also last method is by taking integral have a good method, in this question this algorithm is zhao which is usfull one

Hi!

I am using a proceure to conpute de integral of a function by he Simpson's rule. My function is defined from a function and a procedure, but I am getting the error  "Error, (in w) invalid input: hfun2 expects its 1st argument, t, to be of type numeric, but received (1/10)*i+1/20"

As you can see in the attaxhed file, I have tried several ways to compute the integral but always returns the above error. Please, can yo help me?

Thanks

forum.mw

I upload picture of solution and i try to solve but i fail i don't know how maple can do that just take laplace of one sidehow posible ?


 

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(inttrans)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

eq := diff(y(x), `$`(x, 2))+(1+x)*(diff(y(x), x))-y(x) = 2*y(x)^3

diff(diff(y(x), x), x)+(1+x)*(diff(y(x), x))-y(x) = 2*y(x)^3

(2)

eq1 := laplace(eq, x, s)

s^2*laplace(y(x), x, s)-(D(y))(0)-s*y(0)-2*laplace(y(x), x, s)+s*laplace(y(x)*x, x, s)+s*laplace(y(x), x, s)-y(0) = 2*laplace(y(x)^3, x, s)

(3)

NULL

NULL

NULL

NULL

NULL

NULL

NULL

NULL

NULL

restart

with(inttrans)

[addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace, invmellin, laplace, mellin, savetable, setup]

(4)

eq := diff(y(x, t), x)+y(x, t)^2 = 1

diff(y(x, t), x)+y(x, t)^2 = 1

(5)

eq1 := laplace(eq, x, s)

s*laplace(y(x, t), x, s)-y(0, t)+laplace(y(x, t)^2, x, s) = 1/s

(6)

eq2 := subs(y(0, t) = 3, eq1)

s*laplace(y(x, t), x, s)-3+laplace(y(x, t)^2, x, s) = 1/s

(7)

lap := s*laplace(y(x, t), x, s) = 1/s+3-laplace(y(x, t)^2, x, s)

s*laplace(y(x, t), x, s) = 1/s+3-laplace(y(x, t)^2, x, s)

(8)

eq3 := lap/s

laplace(y(x, t), x, s) = (1/s+3-laplace(y(x, t)^2, x, s))/s

(9)

expand(%)

laplace(y(x, t), x, s) = 1/s^2+3/s-laplace(y(x, t)^2, x, s)/s

(10)

Geq := y(x, t) = invlaplace(1/s^2+3/s, s, x)-invlaplace(laplace(y(x, t)^2, x, s)/s, x, s)

y(x, t) = x+3-invlaplace(laplace(y(x, t)^2, x, s), x, s)/s

(11)

NULL

k := K

K

(12)

f := proc (y) options operator, arrow; y^2 end proc

proc (y) options operator, arrow; y^2 end proc

(13)

for j from 0 to 4 do A[j] := subs(lambda = 0, (diff(f(sum(lambda^i*y[i](x), i = 0 .. 20)), [`$`(lambda, j)]))/factorial(j)) end do

y[0](x)^2

 

2*y[0](x)*y[1](x)

 

y[1](x)^2+2*y[0](x)*y[2](x)

 

2*y[1](x)*y[2](x)+2*y[0](x)*y[3](x)

 

y[2](x)^2+2*y[1](x)*y[3](x)+2*y[0](x)*y[4](x)

(14)

" y[0](x):=x+3"

proc (x) options operator, arrow, function_assign; 3+x end proc

(15)

lapy[1] := -laplace(A[0], x, s)/s

-(9*s^2+6*s+2)/s^4

(16)

simplify(%)

(-9*s^2-6*s-2)/s^4

(17)

invlaplace(%, s, x)

-(1/3)*x*(x^2+9*x+27)

(18)

simplify(-(1/3)*x*(x^2+9*x+27))

-(1/3)*x*(x^2+9*x+27)

(19)

normal(-(1/3)*x*(x^2+9*x+27), ':-expanded')

-(1/3)*x^3-3*x^2-9*x

(20)

"y[1](x):=-1/3 x^3-3 x^2-9 x"

proc (x) options operator, arrow, function_assign; -(1/3)*x^3-3*x^2-9*x end proc

(21)

lapy[2] := -laplace(A[1], x, s)/s

-(-72/s^3-48/s^4-16/s^5-54/s^2)/s

(22)

simplify(-(-72/s^3-48/s^4-16/s^5-54/s^2)/s)

(54*s^3+72*s^2+48*s+16)/s^6

(23)

expand(%)

54/s^3+72/s^4+48/s^5+16/s^6

(24)

invlaplace(%, s, x)

(1/15)*x^2*(2*x^3+30*x^2+180*x+405)

(25)

expand(%)

(2/15)*x^5+2*x^4+12*x^3+27*x^2

(26)

NULL

NULL


 

Download laplace_of_special_ode.mw

example 1 i could't solve but example to i did

i am not jobles to post in here and you delete them one by one, when i post i am waiting for response not you delete them, with this cuase my question are repeat or something like that 
this time delete my post i will upload 1000 of nonsense question and post  and go delete all of them ...

Hi,
I have an equation and I want to solve it parametrically to find x , but I couldn't do that with "solve" command. (I know x should be  real and positive). What should I do?
Root_of.mw

1 2 3 4 5 6 7 Last Page 1 of 2369