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Question Number 132398 by bramlexs22 last updated on 14/Feb/21

$$\mathrm{find}\mathrm{general}\mathrm{solution}{\mathrm{y}}^{\prime }={(64\mathrm{x}+\mathrm{y})}^2$$

Answered by EDWIN88 last updated on 14/Feb/21

$$\mathrm{let}\mathrm{v}=64\mathrm{x}+\mathrm{y}\iff \frac{\mathrm{dv}}{\mathrm{dx}}=64+\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\mathrm{or}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}-64$$$$\iff \frac{\mathrm{dv}}{\mathrm{dx}}-64={\mathrm{v}}^2;\frac{\mathrm{dv}}{\mathrm{dx}}={\mathrm{v}}^2+64$$$$\int \frac{\mathrm{dv}}{{\mathrm{v}}^2+8^2}=\int \mathrm{dx}$$$$\arctan \left(\frac{\mathrm{v}}{8}\right)=8\mathrm{x}+\mathrm{C}$$$$\frac{64\mathrm{x}+\mathrm{y}}{8}=\tan (8\mathrm{x}+\mathrm{C});\mathrm{y}=8\tan (8\mathrm{x}+\mathrm{C})-64\mathrm{x}$$

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