f(x)={0,−π

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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​∫−xπ​f(x)dx =π1​[∫−x0​f(x)dx+∫0pt​f(x)dx] −r21​[∫−π0​x1​∫02​(x+1)dx] =321​[0+(23x2​+2x)∣]] =π1​[23​(x2−02)+2(π−∞)] −x1​[23π2+4x​]=3π+2⇒[a0​=3x+π] ax−21​∫−xx​f(x)cosxdx <xt1​[∫0−x0​0cmixdx+∫ax​(3x+e)cosxdx] =π1​[∫0⋅dx+∫0x​(3x+2)cosxd]=π1​[0+∫0x​(3x+2)cosndx]​u=du=​ =x1​[(3x+2)∫0x​cosndx−∫0x​(dxd​(3x+2)∫con2ndx) =x1​{[(3x+2)h3n+x​]0π​−∫0n​343sindx​d =r1​[0+3n2cos3​] ∵chnA=(−1)
lon=521​∫−2π​sensinxdx =π1​[∫−20​0sindxdx+∫0π​(3x+2)sincosdx] −π1​[0+(3x+2)(n−cosn​)∣∣​0π​−∫0π​n(−cosα)​dx] =π1​[(3x+2)(n−cosax​)+[3n2sinx​]02​1 =π1​[(−3x+2)n(−1)(−1)2​+3(n2sinnπ​) he​=π1​(3x+2)(−1)−4 f(x)=23π+2​+∑π ne 3(−1)n​canux+∑x​3(x+2)c−1)n+1​gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2​yp=? S⋅In=y=ae1+b2tVy(0)=0
Ff(x)=2a​+∑arhx+∑lnsinnx ax=π1​∫−xπ​f(x)dx =π1​[∫−x0​f(x)dx+∫0pt​f(x)dx] −r21​[∫−π0​x1​∫02​(x+1)dx] =321​[0+(23x2​+2x)∣]] =π1​[23​(x2−02)+2(π−∞)] −x1​[23π2+4x​]=3π+2⇒[a0​=3x+π] ax−21​∫−xx​f(x)cosxdx <xt1​[∫0−x0​0cmixdx+∫ax​(3x+e)cosxdx] =π1​[∫0⋅dx+∫0x​(3x+2)cosxd]=π1​[0+∫0x​(3x+2)cosndx]​u=du=​ =x1​[(3x+2)∫0x​cosndx−∫0x​(dxd​(3x+2)∫con2ndx) =x1​{[(3x+2)h3n+x​]0π​−∫0n​343sindx​d =r1​[0+3n2cos3​] ∵chnA=(−1)
lon=521​∫−2π​sensinxdx =π1​[∫−20​0sindxdx+∫0π​(3x+2)sincosdx] −π1​[0+(3x+2)(n−cosn​)∣∣​0π​−∫0π​n(−cosα)​dx] =π1​[(3x+2)(n−cosax​)+[3n2sinx​]02​1 =π1​[(−3x+2)n(−1)(−1)2​+3(n2sinnπ​) he​=π1​(3x+2)(−1)−4 f(x)=23π+2​+∑π ne 3(−1)n​canux+∑x​3(x+2)c−1)n+1​gn
y′−5y+6y=e5,y(0)=0,y(0)=0 A⋅E⋅y′−5y+6y)=0(D−3)(D−2)=0D=3,D=2​yp=? 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.