Solve the following system: x' = 4x - y + t^2 y' = x + 2y + 3t

Question

Solve the following system:

[latex]\begin{aligned}&x^{\prime}=4 x-y+t^{2} \&y^{\prime}=x+2 y+3 t\end{aligned}\quad\quad\quad\quad\quad (12)[/latex]

Accepted Answer

Proceeding as before, we obtain

[latex]\begin{aligned}x^{\prime \prime} &=4 x^{\prime}-y^{\prime}+2 t=4 x^{\prime}-\overbrace{(x+2 y+3 t)}^{y^{\prime}}+2 t \&=4 x^{\prime}-x-2 y-t\end{aligned}[/latex]

From the first equation of [latex](12),-y=x^{\prime}-4 x-t^{2}[/latex], so [latex]-2 y=2 x^{\prime}-8 x-2 t^{2}[/latex] and

[latex]x^{\prime \prime}=4 x^{\prime}-x+\left(2 x^{\prime}-8 x-2 t^{2} ight)-t[/latex]

or, simplifying,

[latex]x^{\prime \prime}-6 x^{\prime}+9 x=-2 t^{2}-t .\quad\quad\quad\quad\quad (13)[/latex]

The associated homogeneous equation is

[latex]x^{\prime \prime}-6 x^{\prime}+9 x=0[/latex]

with characteristic equation

[latex]0=\lambda^{2}-6 \lambda+9=(\lambda-3)^{2}[/latex].

Thus the solution to the homogeneous equation is [latex]x(t)=c_{1} e^{3 t}+c_{2} t e^{3 t}[/latex]. Using the method of undetermined coefficients (Section 11.10), we obtain the particular solution

[latex]-\frac{2}{9} t^{2}-\frac{11}{27} t-\frac{2}{9}[/latex].

Hence the general solution to equation (13) is

[latex]x(t)=c_{1} e^{3 t}+c_{2} t e^{3 x}-\frac{2}{9} t^{2}-\frac{11}{27} t-\frac{2}{9}[/latex].

Also,

[latex]x^{\prime}(t)=3 c_{1} e^{3 t}+3 c_{2} t e^{3 t}+c_{2} e^{3 t}-\frac{4}{9} t-\frac{11}{27}[/latex].

From the first equation in (12),

[latex]y(t)=4 x-x^{\prime}+t^{2},[/latex]

and we obtain

[latex]y(t)=c_{1} e^{3 t}+c_{2} t e^{3 t}-c_{2} e^{3 t}+\frac{1}{9} t^{2}-\frac{32}{27} t-\frac{13}{27}[/latex].

Question 12.1.4: Solve the following system: x' = 4x - y + t^2 y' = x + 2y +...

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