Question #77428

d^2y/dx^2+3(dy/dx)+ 2y=1+3x+x^2

Expert's answer

Answer on Question #77428 – Math – Differential Equations

Question


d2ydx2+3dydx+2y=1+3x+x2.\frac {d ^ {2} y}{d x ^ {2}} + 3 \frac {d y}{d x} + 2 y = 1 + 3 x + x ^ {2}.


Solution

The corresponding homogeneous equation is d2ydx2+3dydx+2y=0\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0 . So, characteristic equation will be


r2+3r+2=(r+1)(r+2)=0.r ^ {2} + 3 r + 2 = (r + 1) (r + 2) = 0.


Then the complementary solution will be


yc=C1ex+C2e2x.y _ {c} = C _ {1} e ^ {- x} + C _ {2} e ^ {- 2 x}.


The nonhomogeneous equation has f(x)=1+3x+x2f(x) = 1 + 3x + x^2 . We will search a particular solution in the next general quadratic polynomial form:


yp=A+Bx+Cx2.y _ {p} = A + B x + C x ^ {2}.


Then dydx=B+2Cx\frac{dy}{dx} = B + 2Cx and d2ydx2=2C\frac{d^2y}{dx^2} = 2C . Substitute them into the equation:


2C+3(B+2Cx)+2(A+Bx+Cx2)=1+3x+x2.2 C + 3 (B + 2 C x) + 2 (A + B x + C x ^ {2}) = 1 + 3 x + x ^ {2}.


The corresponding terms on both sides should have the same coefficients. Hence, we obtain:


2C=1,2 C = 1,6C+2B=3,6 C + 2 B = 3,2C+3B+2A=1.2 C + 3 B + 2 A = 1.


Now we have the next solution for unknown coefficients:


C=12,C = \frac {1}{2},B=0,B = 0,A=0.A = 0.


The general solution of equation is


y=yc+yp=C1ex+C2e2x+x22.y = y _ {c} + y _ {p} = C _ {1} e ^ {- x} + C _ {2} e ^ {- 2 x} + \frac {x ^ {2}}{2}.


Answer: y=C1ex+C2e2x+x22y = C_1 e^{-x} + C_2 e^{-2x} + \frac{x^2}{2} .

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