2.4 Equations reducible to homogeneous form

1.1 Order and degree of a differential equation
1.2 Formation of a differential equation

2.1 Variable separable method
2.2 Equations reducible to variable separable form
2.3 Homogeneous differential equation
2.4 Equations reducible to homogeneous form
2.5 Exact differential equation
2.6 Equations reducible to exact form
2.7 Linear differential equation
2.8 Equations reducible to linear form

(i) A differential equation of the form $$\frac{{dy}}{{dx}} = f\left( {\frac{{{a_1}x + {b_1}y + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}}} \right)$$ where ${a_1}{b_2} - {a_2}{b_1} \ne 0$, then substitute $x=X+h$ and $y=Y+k$ to transform the above equation to a homogeneous type in new variables $X$ and $Y$ i.e., $$\frac{{dY}}{{dX}} = \frac{{{a_1}X + {b_1}Y + ({a_1}h + {b_1}k + {c_1})}}{{{a_2}X + {b_2}Y + ({a_2}h + {b_2}k + {c_2})}}$$

These constants are chosen in such a way that ${a_1}h + {b_1}k + {c_1} = 0$ and ${a_2}h + {b_2}k + {c_2} = 0$. Therefore, the equation becomes $$\frac{{dY}}{{dX}} = \frac{{{a_1}X + {b_1}Y}}{{{a_2}X + {b_2}Y}}$$

which is a homogeneous differential equation and can again be solved by putting $Y=vX$ or $X=vY$ and using variable separable method as suggested earlier in the above method.

Question 8. Solve the differential equation $$(2x - y + 4)dy + (x - 2y + 5)dx = 0$$

Solution: The given differential equation is non-homogeneous and ${a_1}{b_2} - {a_2}{b_1} = (2 \times - 2) - (1 \times - 1) = - 4 + 1 = - 3 \ne 0$

Now, $h$ and $k$ can be find out using $2h-k+4=0$ and $h-2k+5=0$.

On solving them, we get $h=-1$ and $k=2$.

Now, substitute $x=X-1$ and $y=Y+2$ in given differential equation, we get $$(2X - Y)dY + (X - 2Y)dX = 0...(1)$$

which is a homogeneous equation in $X$ and $Y$. Now, put $X = vY \Rightarrow dX = vdY + Ydv$ in $(1)$,

we get

$$ (2vY - Y)dY + (vY - 2Y)(vdY + Ydv) = 0 $$ $$ \Rightarrow (2v - 1)dY + (v - 2)vdv + Y(v - 2)dv = 0 $$ $$ \Rightarrow ({v^2} - 2v + 2v - 1)dY + Y(v - 2)dv = 0 $$ $$ \Rightarrow \int {\frac{{dY}}{Y} + \int {\frac{{v - 2}}{{{v^2} - 1}}dv} } = C $$ $$ \Rightarrow \ln \left| Y \right| + \int {\left\{ { - \frac{1}{{2(v - 1)}} + \frac{3}{{2(v + 1)}}} \right\}dv} = C $$ $$ \ln \left| Y \right| - \frac{1}{2}\ln \left| {v - 1} \right| + \frac{3}{2}\ln \left| {v + 1} \right| = \ln {C_1}\quad [{\text{Assume }}C = \ln {C_1}({C_1} > 0)] $$

Put $v = \frac{X}{Y}$, we get

$$ \ln \left| Y \right| - \frac{1}{2}\ln \left| {\frac{{X - Y}}{Y}} \right| + \frac{3}{2}\ln \left| {\frac{{X + Y}}{Y}} \right| = \ln {C_1} $$ $$ \Rightarrow \ln \left[ {\frac{{\left| Y \right|\left| {{Y^{\frac{1}{2}}}} \right|.{{(X + Y)}^{\frac{3}{2}}}}}{{{{(X - Y)}^{\frac{1}{2}}}.{Y^{\frac{3}{2}}}}}} \right] = \ln {C_1} $$ $$ \therefore {(X + Y)^{\frac{3}{2}}} = {C_1}{(X - Y)^{\frac{1}{2}}} $$ $$ {(X + Y)^3} = {C_2}(X - Y)\quad ({\text{where }}{C_2} = {C_1}^2) $$ $$ {(x + y - 1)^3} = {C_2}(x - y + 3) $$

(ii) A differential equation of the form $$\frac{{dy}}{{dx}} = f\left( {\frac{{{a_1}x + {b_1}y + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}}} \right)$$ where $$\begin{equation} \begin{aligned} {a_1}{b_2} - {a_2}{b_1} = 0 \\ \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \\\end{aligned} \end{equation} $$ then substitute ${a_1}x + {b_1}y = v$ which transforms the given differential equation to an equation in which variable separable method can be applied.