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Question Number 88492 by jagoll last updated on 11/Apr/20

$${\mathrm{y}}^{″}-{4\mathrm{y}}^{\prime }+5\mathrm{y}=1+8\cos \mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}$$

Commented by niroj last updated on 11/Apr/20

$$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}-4\frac{\mathbf{dy}}{\mathbf{dx}}+5\mathbf{y}=1+8\mathbf{cos}\mathbf{x}+{\mathbf{e}}^{2\mathbf{x}}$$$$({\mathrm{D}}^2-4\mathrm{D}+5)\mathrm{y}=1+8\cos \mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}$$$$\mathrm{Auxilairy}\mathrm{Equation},$$$${\mathrm{m}}^2-4\mathrm{m}+5=0$$$$\mathrm{m}=\frac{-(-4)\stackrel{-}{+}\sqrt{{(-4)}^2-4.1.5}}{2.1}$$$$\mathrm{m}=\frac{4\stackrel{-}{+}\sqrt{16-20}}{2}=\frac{4\stackrel{-}{+}\sqrt{-4}}{2}$$$$=2\stackrel{-}{+}\frac{2i}{2}=2\stackrel{-}{+}1i$$$$\mathrm{CF}={\mathrm{e}}^{2\mathrm{x}}({\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\sin \mathrm{x})$$$$\mathrm{PI}=\frac{1+8\cos \mathrm{x}+{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{D}}^2-4\mathrm{D}+5}$$$$=\frac{1}{{\mathrm{D}}^2-4\mathrm{D}+5}+\frac{8\cos \mathrm{x}}{{\mathrm{D}}^2-4\mathrm{D}+5}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{D}}^2-4\mathrm{D}+5}$$$$=\frac{{\mathrm{e}}^{0.\mathrm{x}}}{{\mathrm{D}}^2-4\mathrm{D}+5}+\frac{8\cos \mathrm{x}}{-1-4\mathrm{D}+5}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{D}}^2-4\mathrm{D}+5}$$$$=\frac{{\mathrm{e}}^{0.\mathrm{x}}}{{\left(0\right)}^2-4.0+5}+\frac{8\cos \mathrm{x}}{-4\mathrm{D}+4}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\left(2\right)}^2-4.2+5}$$$$=\frac{1}{5}-\frac{8}{4}.\frac{\cos \mathrm{x}}{(\mathrm{D}-1)}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{4-8+5}$$$$=\frac{1}{5}-\frac{2(\mathrm{D}+1)\cos \mathrm{x}}{({\mathrm{D}}^2-1)}+\frac{{\mathrm{e}}^{2\mathrm{x}}}{1}$$$$=\frac{1}{5}-\frac{2(\mathrm{D}+1)\mathrm{x}\cos \mathrm{x}}{(-1-1)}+{\mathrm{e}}^{2\mathrm{x}}$$$$=\frac{1}{5}-\frac{2(\mathrm{D}+1)\cos \mathrm{x}}{-2}+{\mathrm{e}}^{2\mathrm{x}}$$$$=\frac{1}{5}+\frac{(\mathrm{D}+1)\cos \mathrm{x}}{1}+{\mathrm{e}}^{2\mathrm{x}}$$$$=\frac{1}{5}+(-\sin \mathrm{x}+\cos \mathrm{x})+{\mathrm{e}}^{2\mathrm{x}}$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\mathrm{y}={\mathrm{e}}^{2\mathrm{x}}({\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\mathrm{sinx})+\frac{1}{5}-\sin \mathrm{x}+\mathrm{cosx}+{\mathrm{e}}^{2\mathrm{x}}.$$

Commented by jagoll last updated on 11/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by peter frank last updated on 11/Apr/20

$$thankyou$$

Commented by peter frank last updated on 11/Apr/20

$$thankyou$$

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