f(x)={0,−π
Posted: Thu Jul 14, 2022 4:22 pm
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, I have a mistake by red
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f(x)={0,−π<x≤03x+2,0<x≤π interval (−π,π). 5. Find the solutions of the given differential equations with the given conditions y′′−5y′+6y=e5t,y(0)=0,y′(0)=0.
Ff(x)=2a+∑arhx+∑lnsinnx ax=π1∫−xπf(x)dx =π1[∫−x0f(x)dx+∫0ptf(x)dx] −r21[∫−π0x1∫02(x+1)dx] =321[0+(23x2+2x)∣]] =π1[23(x2−02)+2(π−∞)] −x1[23π2+4x]=3π+2⇒[a0=3x+π] ax−21∫−xxf(x)cosxdx <xt1[∫0−x00cmixdx+∫ax(3x+e)cosxdx] =π1[∫0⋅dx+∫0x(3x+2)cosxd]=π1[0+∫0x(3x+2)cosndx]u=du= =x1[(3x+2)∫0xcosndx−∫0x(dxd(3x+2)∫con2ndx) =x1{[(3x+2)h3n+x]0π−∫0n343sindxd =r1[0+3n2cos3] ∵chnA=(−1)
lon=521∫−2πsensinxdx =π1[∫−200sindxdx+∫0π(3x+2)sincosdx] −π1[0+(3x+2)(n−cosn)∣∣0π−∫0πn(−cosα)dx] =π1[(3x+2)(n−cosax)+[3n2sinx]021 =π1[(−3x+2)n(−1)(−1)2+3(n2sinnπ) he=π1(3x+2)(−1)−4 f(x)=23π+2+∑π ne 3(−1)ncanux+∑x3(x+2)c−1)n+1gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2yp=? S⋅In=y=ae1+b2tVy(0)=0
Ff(x)=2a+∑arhx+∑lnsinnx ax=π1∫−xπf(x)dx =π1[∫−x0f(x)dx+∫0ptf(x)dx] −r21[∫−π0x1∫02(x+1)dx] =321[0+(23x2+2x)∣]] =π1[23(x2−02)+2(π−∞)] −x1[23π2+4x]=3π+2⇒[a0=3x+π] ax−21∫−xxf(x)cosxdx <xt1[∫0−x00cmixdx+∫ax(3x+e)cosxdx] =π1[∫0⋅dx+∫0x(3x+2)cosxd]=π1[0+∫0x(3x+2)cosndx]u=du= =x1[(3x+2)∫0xcosndx−∫0x(dxd(3x+2)∫con2ndx) =x1{[(3x+2)h3n+x]0π−∫0n343sindxd =r1[0+3n2cos3] ∵chnA=(−1)
lon=521∫−2πsensinxdx =π1[∫−200sindxdx+∫0π(3x+2)sincosdx] −π1[0+(3x+2)(n−cosn)∣∣0π−∫0πn(−cosα)dx] =π1[(3x+2)(n−cosax)+[3n2sinx]021 =π1[(−3x+2)n(−1)(−1)2+3(n2sinnπ) he=π1(3x+2)(−1)−4 f(x)=23π+2+∑π ne 3(−1)ncanux+∑x3(x+2)c−1)n+1gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2yp=? S⋅In=y=ae1+b2tVy(0)=0
f(x)={0,−π<x≤03x+2,0<x≤π interval (−π,π). 5. Find the solutions of the given differential equations with the given conditions y′′−5y′+6y=e5t,y(0)=0,y′(0)=0.
Posted: Thu Jul 14, 2022 4:22 pm
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Ff(x)=2a+∑arhx+∑lnsinnx ax=π1∫−xπf(x)dx =π1[∫−x0f(x)dx+∫0ptf(x)dx] −r21[∫−π0x1∫02(x+1)dx] =321[0+(23x2+2x)∣]] =π1[23(x2−02)+2(π−∞)] −x1[23π2+4x]=3π+2⇒[a0=3x+π] ax−21∫−xxf(x)cosxdx <xt1[∫0−x00cmixdx+∫ax(3x+e)cosxdx] =π1[∫0⋅dx+∫0x(3x+2)cosxd]=π1[0+∫0x(3x+2)cosndx]u=du= =x1[(3x+2)∫0xcosndx−∫0x(dxd(3x+2)∫con2ndx) =x1{[(3x+2)h3n+x]0π−∫0n343sindxd =r1[0+3n2cos3] ∵chnA=(−1)
lon=521∫−2πsensinxdx =π1[∫−200sindxdx+∫0π(3x+2)sincosdx] −π1[0+(3x+2)(n−cosn)∣∣0π−∫0πn(−cosα)dx] =π1[(3x+2)(n−cosax)+[3n2sinx]021 =π1[(−3x+2)n(−1)(−1)2+3(n2sinnπ) he=π1(3x+2)(−1)−4 f(x)=23π+2+∑π ne 3(−1)ncanux+∑x3(x+2)c−1)n+1gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2yp=? S⋅In=y=ae1+b2tVy(0)=0
Ff(x)=2a+∑arhx+∑lnsinnx ax=π1∫−xπf(x)dx =π1[∫−x0f(x)dx+∫0ptf(x)dx] −r21[∫−π0x1∫02(x+1)dx] =321[0+(23x2+2x)∣]] =π1[23(x2−02)+2(π−∞)] −x1[23π2+4x]=3π+2⇒[a0=3x+π] ax−21∫−xxf(x)cosxdx <xt1[∫0−x00cmixdx+∫ax(3x+e)cosxdx] =π1[∫0⋅dx+∫0x(3x+2)cosxd]=π1[0+∫0x(3x+2)cosndx]u=du= =x1[(3x+2)∫0xcosndx−∫0x(dxd(3x+2)∫con2ndx) =x1{[(3x+2)h3n+x]0π−∫0n343sindxd =r1[0+3n2cos3] ∵chnA=(−1)
lon=521∫−2πsensinxdx =π1[∫−200sindxdx+∫0π(3x+2)sincosdx] −π1[0+(3x+2)(n−cosn)∣∣0π−∫0πn(−cosα)dx] =π1[(3x+2)(n−cosax)+[3n2sinx]021 =π1[(−3x+2)n(−1)(−1)2+3(n2sinnπ) he=π1(3x+2)(−1)−4 f(x)=23π+2+∑π ne 3(−1)ncanux+∑x3(x+2)c−1)n+1gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2yp=? S⋅In=y=ae1+b2tVy(0)=0
f(x)={0,−π<x≤03x+2,0<x≤π interval (−π,π). 5. Find the solutions of the given differential equations with the given conditions y′′−5y′+6y=e5t,y(0)=0,y′(0)=0.