Question #339562

Find the general solution of the following differential equation using the method of



undetermined coefficients: d^y/dx^2 - dy/dx + 2y = 2cosh2x

1
Expert's answer
2022-05-11T15:27:52-0400

At first, we solve the homogeneous equation: gg+2g=0g''-g'+2g=0. The characteristic equation has the form: λ2λ+2=0\lambda^2-\lambda+2=0. The solution is: λ1=1i72\lambda_1=\frac{1-i\sqrt{7}}{2}, λ2=1+i72\lambda_2=\frac{1+i\sqrt{7}}{2}. Thus, g=C1eλ1x+C2eλ2xg=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}with C1,C2CC_1,C_2\in{\mathbb{C}} is the solution of equation: gg+2g=0g''-g'+2g=0. The solution can be rewritten in the following form: g=k1e12xsin(72x)+k2e12xcos(72x)g=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x) with k1,k2Ck_1,k_2\in{\mathbb{C}}. We will try to find a solution of inhomogeneous equation in the form: f=aeαx+beβxf=ae^{\alpha x}+be^{\beta x} with a,b,α,βCa,b,\alpha,\beta\in{\mathbb{C}} . We receive: f=aαeαx+bβeβxf'=a\alpha e^{\alpha x}+b\beta e^{\beta x}, f=aα2eαx+bβ2eβxf''=a\alpha^2e^{\alpha x}+b\beta^2e^{\beta x}. ff+2f=a(α2α+2)eαx+b(β2β+2)eβxf''-f'+2f=a(\alpha^2-\alpha+2)e^{\alpha x}+b(\beta^2-\beta+2)e^{\beta x} . Right side of equation has the form: e2x+e2xe^{2x}+e^{-2x}. We put α=2\alpha=2, β=2\beta=-2 and receive: ff+2f=4ae2x+8be2x=e2x+e2xf''-f'+2f=4ae^{2x}+8be^{-2x}=e^{2x}+e^{-2x}. From the latter we get: a=14a=\frac14, b=18b=\frac18. A solution is f=14e2x+18e2xf=\frac14e^{2x}+\frac18e^{-2x}. The general solution of the initial equation is: y=f+g=k1e12xsin(72x)+k2e12xcos(72x)+14e2x+18e2xy=f+g=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x)+\frac14e^{2x}+\frac18e^{-2x} with k1,k2Ck_1,k_2\in{\mathbb{C}}.

Answer: the solution is: y=k1e12xsin(72x)+k2e12xcos(72x)+14e2x+18e2xy=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x)+\frac14e^{2x}+\frac18e^{-2x} with k1,k2Ck_1,k_2\in{\mathbb{C}}.


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