Homogeneous differential equation

Question

Solve the differential equation

1.y'=(y-x)^2

2.y'=tan(x+y) - 1

Answer 1

(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

\(\Rightarrow\) \(\frac{v-1}{v+1}\) = ce^x

\(\Rightarrow\) 1 - \(\frac{2}{v+1}\) = ce^x

\(\Rightarrow\) \(\frac{2}{v+1}\) = 1 - ce^x

\(\Rightarrow\) 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 = ce^x 

\(\Rightarrow\) sin(x + y ) = ce^x