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Question Number 178415 by Spillover last updated on 16/Oct/22

$$\mathrm{If}\mathrm{acosh}\mathrm{x}+\mathrm{bsinh}\mathrm{x}=\mathrm{c}$$$$\mathrm{show}\mathrm{that}.$$$$\mathrm{x}=\ln \left[\frac{\mathrm{c}\pm \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2}}{\mathrm{a}+\mathrm{b}}\right]$$

Answered by haladu last updated on 16/Oct/22

$$$$$$\mathbf{a}\cosh \mathbf{x}+\mathbf{b}\sinh \mathbf{x}=\mathbf{C}$$$$$$$$\cosh \left(\mathbf{x}\right)=\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}}{2}$$$$\sinh \left(\mathbf{x}\right)=\frac{{\mathbf{e}}^{\mathbf{x}}-{\mathbf{e}}^{-\mathbf{x}}}{2}$$$$$$$$\mathbf{a}\left\{\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}}{2}\right\}+\mathbf{b}\left\{\frac{{\mathbf{e}}^{\mathbf{x}}-{\mathbf{e}}^{-\mathbf{x}}}{2}\right\}=\mathbf{c}$$$$$$$$\mathbf{mutiply}\mathbf{both}\mathbf{sides}\mathbf{by}{\mathbf{e}}^{\mathbf{x}}$$$$$$$$\frac{\mathbf{a}}{2}\{{\left({\mathbf{e}}^{\mathbf{x}}\right)}^2+1\}+\frac{\mathbf{b}}{2}\{{\left({\mathbf{e}}^{\mathbf{x}}\right)}^2-1\}={\mathbf{ce}}^{\mathbf{x}}$$$$$$$${\left({\mathbf{e}}^{\mathbf{x}}\right)}^2\left(\frac{\mathbf{a}+\mathbf{b}}{2}\right)+\frac{\mathbf{a}-\mathbf{b}}{2}={\mathbf{ce}}^{\mathbf{x}}$$$$$$$${\left({\mathbf{e}}^{\mathbf{x}}\right)}^2\left(\frac{\mathbf{a}+\mathbf{b}}{2}\right)-{\mathbf{ce}}^{\mathbf{x}}+\frac{\mathbf{a}-\mathbf{b}}{2}=0$$$$$$$$\mathbf{mutiply}\mathbf{both}\mathbf{sides}\mathbf{by}\frac{2}{\mathbf{a}+\mathbf{b}}$$$$$$$$\Rightarrow {\left({\mathbf{e}}^{\mathbf{x}}\right)}^2-\frac{2\mathbf{c}}{\mathbf{a}+\mathbf{b}}\left({\mathbf{e}}^{\mathbf{x}}\right)+\frac{\mathbf{a}-\mathbf{b}}{\mathbf{a}+\mathbf{b}}=0$$$$$$$$\Rightarrow {\left({\mathbf{e}}^{\mathbf{x}}\right)}^2+2(-\frac{\mathbf{c}}{\mathbf{a}+\mathbf{b}})\left({\mathbf{e}}^{\mathbf{x}}\right)=\frac{\mathbf{b}-\mathbf{a}}{\mathbf{a}+\mathbf{b}}$$$$$$$$\Rightarrow {({\mathbf{e}}^{\mathbf{x}}-\frac{\mathbf{c}}{\mathbf{a}+\mathbf{b}})}^2=\frac{\mathbf{b}-\mathbf{a}}{\mathbf{a}+\mathbf{b}}+\frac{{\mathbf{c}}^2}{{(\mathbf{a}+\mathbf{b})}^2}$$$$$$$${\mathbf{e}}^{\mathbf{x}}-\frac{\mathbf{c}}{\mathbf{a}+\mathbf{b}}=\sqrt{\frac{(\mathbf{b}-\mathbf{a})(\mathbf{b}+\mathbf{a})+{\mathbf{c}}^2}{{(\mathbf{a}+\mathbf{b})}^2}}$$$$$$$$\Rightarrow {\mathbf{e}}^{\mathbf{x}}=\frac{-\mathbf{c}\pm \sqrt{{\mathbf{b}}^2-{\mathbf{a}}^2+{\mathbf{c}}^2}}{(\mathbf{a}+\mathbf{b})}$$$$$$$$\Rightarrow \mathbf{x}=\ln \left\{\frac{-\mathbf{c}\pm \sqrt{{\mathbf{b}}^2+{\mathbf{c}}^2-{\mathbf{a}}^2}}{\mathbf{a}+\mathbf{b}}\right\}$$

Commented by Spillover last updated on 16/Oct/22

$$thankyou$$

Commented by haladu last updated on 16/Oct/22

$$$$$$\mathbf{You}\mathbf{are}\mathbf{wellcome}$$

Answered by CElcedricjunior last updated on 16/Oct/22

$$\mathbf{on}\mathbf{acoshx}+\mathbf{bsinhx}=\mathbf{c}$$$$=>\mathbf{acoshx}+\mathbf{bsinhx}=\mathbf{a}\left(\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}}{2}\right)+\mathbf{b}\left(\frac{{\mathbf{e}}^{\mathbf{x}}-{\mathbf{e}}^{-\mathbf{x}}}{2}\right)$$$$=$$$$=>(\mathbf{a}+\mathbf{b}){\mathbf{e}}^{2\mathbf{x}}+(\mathbf{a}-\mathbf{b})-2{\mathbf{ce}}^{\mathbf{x}}=0$$$$=>\mathrm{\Delta }=4{\mathbf{c}}^2-4({\mathbf{a}}^2-{\mathbf{b}}^2)$$$$\mathbf{supposons}\mathbf{que}\mathbf{\Delta }>0\mathbf{ie}{\mathbf{c}}^2>{\mathbf{a}}^2-{\mathbf{b}}^2=>\mathbf{\Delta }=2\sqrt{{\mathbf{c}}^2+{\mathbf{b}}^2-{\mathbf{a}}^2}$$$$=>{\mathbf{e}}^{\mathbf{x}}=\frac{\mathbf{c}\mp \sqrt{{\mathbf{c}}^2+{\mathbf{b}}^2-{\mathbf{a}}^2}}{\mathbf{a}+\mathbf{b}}$$$$=>\mathbf{x}=\mathbf{ln}\mid \frac{\mathbf{c}\mp \sqrt{{\mathbf{c}}^2+{\mathbf{b}}^2-\mathbf{c}}}{\mathbf{a}+\mathbf{b}}\mid$$$$$$$$............lecelebrecedricjunior.......$$

Commented by Spillover last updated on 16/Oct/22

$$thanks$$

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