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These are questions asked by arshl

Hi. I want to take Laplace transform from differential equation with the final condition(tf).   If the differential equations have initial conditions  it appears in Laplace (for example laplace transform of diff(u,t) is U-s*u(0) that u(0) is initial condition)but for final conditions or boundry conditions What to do ?my equation is



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



u is function of x,t and u(x(10),10)=0

u[1] := (1/2)*x^2*t; L1 := laplace(eq1, t, s)





u(x(10),10)=0; desired output u[1]=x^2(t-10)/2

s1 := u[1]; N[1] := (1/2)*(diff(s1, x))^2+x*(diff(s1, x)); ul[2] := laplace(N[1], t, s); u[2] := invlaplace(ul[2]/s, s, t)









I want to apply: u[2](x(10),10)=0





Hi. I want to get a Fourier transform under the equation How to do this? Equation_is.docx

Hi. what is reason maple unable to integral in answer of dsolve? when I try to use dsolve for solve my equation in answer of maple there are expressions of integral that isnot calculated

∫(-625 R^2 ro (-2 cp3 x1 lambdaopt+sin((pi t)/10) cp2+(16 cp2)/5) (-cos((pi t)/10)+cos(2 pi)+((-t/5+4) x1+(8 t)/25-32/5) pi) lambdaopt pi b11 (e)^(-((16+5 sin((pi t)/10)) cp1)/(10 lambdaopt x1))+625 (-R^2 k11 (cos((pi t)/10))^3+k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) (cos((pi t)/10))^2+((32 sin((pi t)/10) R^2 k11)/5+(281 R^2 k11)/25+4 lambdaopt^2 (a11 x1+a13 x3)) cos((pi t)/10)-(32 k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) sin((pi t)/10))/5+(281 (x1-8/5) (R^2 k11+(100 lambdaopt^2 (a11 x1+a13 x3))/281) pi t)/125) x1^2)ⅆt

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