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 $(\mathrm{h}, \mathrm{k})=$

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Aicharya · Accepted Answer

$\because \frac{d y}{d x}=\frac{2 x+3 y-7}{3 x+2 y-8}$ ...(i) Let us use the transformation : $\begin{aligned} & X=x+h \Rightarrow x=X-h \ & Y=y+k \Rightarrow y=Y-k \ & 	ext { And } \frac{d X}{d Y}=\frac{d x}{d y}\end{aligned}$ 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 $\begin{aligned} & \mathrm{h}=2 \& \mathrm{k}=1 \ & (\mathrm{~h}, \mathrm{k})=(2,1)\end{aligned}$

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…

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