Solution of diffrential equation dy/dx = …
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Posted by Uday Shukla 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
{tex}Put x+y=t \begin{aligned} & \frac{d y}{d x}=\frac{d t}{d x}-1 \\ & \frac{d t}{d x}-1=\operatorname{sect} \\ & \frac{d t}{d x}=\operatorname{sect}+1 \\ & \int \frac{d t}{\operatorname{sect}+1}=\int d x \\ & \int \frac{\operatorname{costdt}}{1+\operatorname{cost}} d t=x \\ & \int \frac{1+\operatorname{costdt}-1}{1+\operatorname{cost}} d t=x \\ & \int\left(d t-\frac{1}{1+\cos t}\right) d t=x \end{aligned} {/tex}
{tex}\begin{aligned} & \mathrm{t}-\frac{1}{2} \int \sec ^2 \frac{\mathrm{t}}{2} \mathrm{dt}=\mathrm{x} \\ & \mathrm{t}-\tan \frac{\mathrm{t}}{2}=\mathrm{x}+\mathrm{C} \\ & \mathrm{x}+\mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\mathrm{x}+\mathrm{C} \\ & \mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=C \end{aligned}{/tex}
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