Question #162123

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Find the particular integral solution of  d^3y/dx^3 + 4(dy/dx)=x+3cosx+ e^-2x


1
Expert's answer
2021-02-16T12:34:23-0500

Solution.

d3ydx3+4dydx=x+3cosx+e2x.\frac{d^3y}{dx^3}+4\frac{dy}{dx}=x+3\cos⁡{x}+e^{−2x}.

This is a linear differential equation of the third-order with constant coefficients. The general solution of this equation is found as the sum of the general solution y~\tilde{y} of the corresponding homogeneous equation and some particular integral solution yy_* of the inhomogeneous equation: y=y~+y.y=\tilde{y}+y_*.

Сompose and solve the characteristic equation:


λ3+4λ=0,λ(λ2+4)=0,λ1=0,or λ2+4=0, λ2=2i, λ3=2i.\lambda^3+4\lambda=0, \newline\lambda(\lambda^2+4)=0, \newline \lambda_1=0, \text{or } \lambda^2+4=0, \newline \text{ } \lambda_2=2i, \text{ } \lambda_3=-2i.


Therefore, y~=C1eλ1x+C2eaxcosbx+C3eaxsinbx,\tilde{y}=C_1e^{\lambda_1x}+C_2e^{ax}\cos{bx}+C_3e^{ax}\sin{bx},

where aa - real part, and bb - imaginary part of a complex number λ2 and λ3.\lambda_2 \text{ and } \lambda_3. So,


y~=C1+C2cos2x+C3sin2x.\newline \tilde{y}=C_1+C_2\cos{2x}+C_3\sin{2x}.

The particular integral solution yy_* write in the form:


y=y1+y2+y3.y_∗=y_1+y_2+y_3​.

Since the right-hand side contains a cosx\cos{x} , we look for a particular integral solution y1y_1 ​in the form:


y1=Acosx+Bsinx.y_1=A\cos{x}+B\sin{x}.

Using the method of indefinite coefficients we find that A=0A=0 and B=1.B=1. So, y1=sinx.y_1=\sin{x}.

Since the right-hand side also contains a e2xe^{-2x} , we find a particular integral solution y2y_2 ​in the form:


y2=Ce2x.y_2=Ce^{-2x}.

Using the method of indefinite coefficients we find that C=116.C=-\frac{1}{16}.

Therefore, y2=116e2x.y_2=-\frac{1}{16}e^{-2x}.

The number is characteristic, so y3y_3 is written as


y3=x(Dx+E).y_3=x(Dx+E).

Using the method of indefinite coefficients we find that D=18D=\frac{1}{8} and E=0.E=0. So, y3=x(18x+0)=18x2.y_3=x(\frac{1}{8}x+0)=\frac{1}{8}x^2.

As follows, the particular integral solution of d3ydx3+4dydx=x+3cosx+e2x\frac{d^3y}{dx3}+4\frac{dy}{dx}=x+3\cos⁡{x}+e^{−2x} is


y=sinx116e2x+18x2,y_*=\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2,

and the general solution of the same equation is

y=C1+C2cos2x+C3sin2x+sinx116e2x+18x2.y=C_1+C_2\cos{2x}+C_3\sin{2x}+\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2.

Answer.

y=sinx116e2x+18x2,y_*=\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2, where yy_* is the particular integral solution.


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