Q119 / 160 · AP EAMCET 2023 (18 May Shift 1) · Mathematics

If $X=x+h, Y=y+k$ transforms $\frac{d y}{d x}=\frac{2 x+3 y-7}{3 x+2 y-8}$ to a homogeneous differential equation, then $(\mathrm{h}, \mathrm{k})=$

  • A
    $(1,2)$
  • Correct answer
    B
    $(2,1)$
  • C
    $(7,8)$
  • D
    $(8,7)$

Solution

$\because \frac{d y}{d x}=\frac{2 x+3 y-7}{3 x+2 y-8}$ ...(i) Let us use the transformation : $\begin{aligned} & X=x+h \Rightarrow x=X-h \\ & Y=y+k \Rightarrow y=Y-k \\ & \text { And } \frac{d X}{d Y}=\frac{d x}{d y}\end{aligned}$ Using above transformation in eq (i), we get $\begin{aligned} & \frac{d Y}{d X}=\frac{2(X-h)+3(Y-k)-7}{3(X-h)+2(Y-k)-8} \\ & \Rightarrow \frac{d Y}{d X}=\frac{2 X+3 Y+(-2 h-3 k-7)}{3 X+2 Y+(-3 h-2 k-8)}\end{aligned}$ To make above differential equation homogeneous: $-2 h-3 k-7=0$ ...(ii) $\&-3 h-2 k-8=0$ ...(iii) Solving equations (ii) \& (iii), we gt $\begin{aligned} & \mathrm{h}=2 \& \mathrm{k}=1 \\ & (\mathrm{~h}, \mathrm{k})=(2,1)\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions