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Question Number 166764 by amin96 last updated on 27/Feb/22

$$\frac{\mathbf{dy}}{\mathbf{dx}}=\mathbf{cos}(\mathbf{y}-\mathbf{x})$$

Answered by mr W last updated on 27/Feb/22

$$letu=y-x$$$$y=u+x$$$$\frac{dy}{dx}=\frac{du}{dx}+1$$$$\frac{du}{dx}+1=\cos u$$$$\frac{du}{\cos u-1}=dx$$$$\int \frac{du}{\cos u-1}=\int dx$$$$\int \frac{d\left(\frac{u}{2}\right)}{-{\sin }^2\frac{u}{2}}=\int dx$$$$\int d\left(\cot \frac{u}{2}\right)=\int dx$$$$\cot \frac{u}{2}=x+C$$$$u=2{\cot }^{-1}(x+C)$$$$y=2{\cot }^{-1}(x+C)+x$$

Commented by amin96 last updated on 27/Feb/22

$$\frac{\mathit{d}\mathit{u}}{\mathit{c}\mathit{o}\mathit{s}\mathit{u}-1}=\mathit{d}\mathit{x}\int \frac{\mathit{d}\mathit{u}}{\mathit{c}\mathit{o}\mathit{s}\mathit{u}-1}=\int \mathit{d}\mathit{x}$$$$\frac{1+\mathit{c}\mathit{o}\mathit{s}\mathit{u}}{\mathit{s}\mathit{i}\mathit{n}\mathit{u}}=\mathit{x}+\mathit{c}\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\left(\mathit{u}\right)+\mathit{c}\mathit{t}\mathit{g}\left(\mathit{u}\right)=\mathit{x}+\mathit{c}$$$$\mathbf{cot}\left(\frac{\mathbf{u}}{2}\right)=\mathbf{x}+\mathbf{c}$$$$\frac{\mathbf{u}}{2}=\mathbf{arccot}(\mathbf{x}+\mathbf{c})\mathbf{u}=2\mathbf{arccot}(\mathbf{x}+\mathbf{c})$$$$\mathbf{y}=2\mathbf{arccot}(\mathbf{x}+\mathbf{c})+\mathbf{x}$$

Commented by amin96 last updated on 27/Feb/22

$$\mathit{y}-\mathit{x}=2\pi \mathit{k}\mathit{k}\in \mathit{Z}itscorrect?$$

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