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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Solve: (D^2-2D+2)Y=Sin2x+5
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Solve: (D^2-2D+2)Y=Sin2x+5

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(D^2-3DD' 2D'^2)Z=SINX COSX
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Home tution for maths and science ( class 8 to 12 ) CBSE ,ICSE Bords only

We have ? = 2 as a root of multiplicity 1, so it contributes the basic solution e2x. ... B. (D ? 1)(D ? 2)(D2 ? 4D + 13)2y = 0. Answer: .... E. (D2 ? 3D + 2)y = x2 sin(2x). ... 6. 25 x ?. 227. 2000. ) + cos (2 x). ( 3. 20 x2 +. 7. 100 x ?. 189. 2000. )]
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doing be at jadavpur university

GO THROUGH LAPLACE TRANSFORM............. THEN (S^2-2S+2)(Y(S))=2/(4+S^2)+5/S NOW DOING PARTIAL FRACTION Y(S)=(1/3)(S/(4+S^2))+(1/6)(2/(4+S^2))-(17/6)((S-1)/((S-1)^2+1)+5/(2S)+(5/2)*(1/(S-1)^2+1)) NOW DO INV LAPLACE TRANSFORM Y=(1/3)COS(2X)+(1/6)SIN(2X)+5/2+(EXP(X))(5/2(SIN X)+(17/6)COS X)
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Experimental Physicist

Solution: y: Exp(x)(c1 cosx+c2sinx)+ cos2x/5-sin2x/10+5/2
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

Though this is not a XII class question. Let me try to explain you. The solution will have Complementary function as well as Particular Integral. (D-1)^2 has repeating roots CF will be (cx+d)e^x and PI can be calculated using standard rules where D^2 will be replaced by -4
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Best Mathematics Tutor in Bhubaneswar

Y=(Sin2X+5)/(D^2-2D+2)
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y=e^x(c1cosx + c2sinx) + 1/10(2cos2x - sin2x) - 5/2
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y= e^x(c1 cosx+c2 sinx) + 1/10 (2cos2x -sin2x ) - 5/2
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The solution has two parts viz general solution yh(x) of associated homogenous equation and particular solution yp(x). y(x) = yh(x) + yp (x) . Solving for yh(x): Writing the characteristic equation for the above differnetial equation . m^2-2m+2 =0 . m = -1+i and -1-i. The solution is...
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The solution has two parts viz general solution yh(x) of associated homogenous equation and particular solution yp(x). y(x) = yh(x) + yp (x) . Solving for yh(x): Writing the characteristic equation for the above differnetial equation . m^2-2m+2 =0 . m = -1+i and -1-i. The solution is yh(x) = e^-x(Acosx+B sinX). Solving for yp(x). yp(x) = 1/(D^2-2D+2) * [sin2x+5] =1/(D^2-2D+2) [sin2x] + 1/(D^2-2D+2) [5] =-1/2D [sin2x] + 5/2 =cos2x/4 + 5/2 There fore the complete solution is yh(x) + yp(x) = e^-x[Acosx+Bsinx] + cos2x/4 +5/2 read less
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