CUET Preparation Today
| Solution of the differential equation \((x-1)dy=ydx,y(0)=-5\) is |
\(y=5(x-1)\) \(y=-5(x-1)\) \(y=-x+1\) \(y=-5(x+1)\) |
| \(y=5(x-1)\) |
| \(\begin{aligned}\frac{dy}{y}&=\frac{dx}{x-1}\\ \text{So, }\log y&=\log(x-1)+\log c\\ y&=c(x-1)\\ \text{Setting }x&=0 \text{ and }y=-5 \text{ we have },\\ c&=5\\ \text{Thus, }y&=5(x-1)\end{aligned}\) |
