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Alborosie
3 years ago
5

Solve the given initial-value problem.

Mathematics
1 answer:
jeka943 years ago
3 0

For the corresponding homogeneous ODE,

y'''-2y''+y'=0

the characteristic equation is

r^3-2r^2+r=r(r-1)^2=0

which admits the characteristic solution,

y_c=C_1+C_2e^x+C_3xe^x

Assume a particular solution of the form

y_p=ax+bx^2e^x+ce^{5x}

(ax because a constant solution is already accounted for by C_1; x^2e^x because both e^x and xe^x are accounted for)

\implies{y_p}'=a+(2x+x^2)be^x+5ce^{5x}

\implies{y_p}''=(2+4x+x^2)be^x+25ce^{5x}

\implies{y_p}'''=(6+6x+x^2)be^x+125ce^{5x}

Substituting the derivatives of y_p into the ODE gives

((6+6x+x^2)be^x+125ce^{5x})-2((2+4x+x^2)be^x+25ce^{5x})+(a+(2x+x^2)be^x+5ce^{5x})=2-24e^x+40e^{5x}

a+2be^x+80ce^{5x}=2-24e^x+40e^{5x}

\implies\begin{cases}a=2\\2b=-24\\80c=40\end{cases}\implies a=2,b=-12,c=\dfrac12

So the particular solution is

y=C_1+C_2e^x+C_3xe^x+2x-12x^2e^x+\dfrac12e^{5x}

With the given initial conditions, we find

y(0)=\dfrac12=C_1+C_2+\dfrac12\implies C_1+C_2=0

y'(0)=\dfrac52=C_2+C_3+2+\dfrac52\implies C_2+C_3=-2

y''(0)=-\dfrac{11}2=C_2+2C_3-6+\dfrac{25}2\implies C_2+2C_3=-12

\implies C_1=10,C_2=-10,C_3=8

and so

\boxed{y(x)=10-10e^x+8xe^x+2x-12x^2e^x+\dfrac12e^{5x}}

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