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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2009 · Mathematics · Three Dimensional Geometry

Solution of the differential equation ( (d y)/(d x)+(y)/(x)= x) is

Solution of the differential equation (dydx+yx=sin⁡\frac{d y}{d x}+\frac{y}{x}=\sinx) is

Options

  1. A(x(y+\cos x)=\cos x+C)
  2. B(x(y-\cos x)=\sin x+C)
  3. C(x(y+\cos x)=\sin x+C)
  4. DNone of these

Correct answer

C. (x(y+\cos x)=\sin x+C)

Step-by-step solution

dydx+1x⋅y=sin⁡x[ Type dydx+Py=Q]e∫Pdx=e∫1xdx=elog⁡x=x\begin{aligned}& \frac{d y}{d x}+\frac{1}{x} \cdot y=\sin x\left[\text{ Type } \frac{d y}{d x}+ P y=Q\right] \\ & e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x \end{aligned}) (\therefore) Sol. is (\mathrm{yx}=\int \mathrm{x} \sin \mathrm{xdx}+\mathrm{C}) =x(−cos⁡x)−∫1⋅(−cos⁡x)dx+C=−xcos⁡x+sin⁡x+C⇒x(y+cos⁡x)=sin⁡x+C\begin{aligned}& =x(-\cos x)-\int1 \cdot(-\cos x) d x+C \\ & =-x \cos x+\sin x+C \Rightarrow x(y+\cos x)=\sin x+C \end{aligned})$
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