(1) y' = (y - x)2
put y - x = v
\(\frac{dy}{d\mathrm{x}}\) - 1 = \(\frac{dv}{d\mathrm{x}}\)
\(\frac{dy}{d\mathrm{x}}\) = 1 + \(\frac{dv}{d\mathrm{x}}\)
1 + \(\frac{dv}{d\mathrm{x}}\) = v2
\(\Rightarrow\) \(\frac{dv}{d\mathrm{x}}\) = v2 - 1
\(\Rightarrow\) \(\frac{dv}{v^2-1}\) = dx
\(\Rightarrow\) log \(\big(\frac{v-1}{v+1}\big)\)= x + c1
\(\Rightarrow\) \(\frac{v-1}{v+1}\) = cex
\(\Rightarrow\) 1 - \(\frac{2}{v+1}\) = cex
\(\Rightarrow\) \(\frac{2}{v+1}\) = 1 - cex
\(\Rightarrow\) v + 1 = \(\frac{2}{1-ce^{\mathrm{x}}}\)
\(\Rightarrow\) v = \(\frac{2}{1-ce^{\mathrm{x}}}\) - 1 = \(\frac{1+ce^{\mathrm{x}}}{1-ce^\mathrm{x}}\)
Now putting the value of v = y - x, we get
y - x = \(\frac{1+ce^{\mathrm{x}}}{1-ce^\mathrm{x}}\)
\(\Rightarrow\) y = x + \(\frac{1+ce^{\mathrm{x}}}{1-ce^\mathrm{x}}\)
(2) y' = tan(x+y)-1
x + y = v
1 + \(\frac{dy}{d\mathrm{x}}\) = \(\frac{dv}{d\mathrm{x}}\)
\(\Rightarrow\) \(\frac{dy}{d\mathrm{x}}\) = \(\frac{dv}{d\mathrm{x}}\) - 1
\(\frac{dv}{d\mathrm{x}}\) - 1 = tan v - 1 \(\Rightarrow\) \(\frac{dv}{d\mathrm{x}}\) = tan v
\(\Rightarrow\) \(\frac{dv}{\tan v}\) = dx
\(\Rightarrow\) \(\int\frac{dv}{\tan v}\) = x
\(\Rightarrow\) \(\int \cot vdv \) = x
\(\Rightarrow\) log sinv = x + c1
\(\Rightarrow\) sinv = cex
\(\Rightarrow\) sin(x + y ) = cex