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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Differential Equations

NDA 2019 · Mathematics · Differential Equations

The solution of the differential equation (d y)/(d x)= (y-x)+1 is

The solution of the differential equation \frac{d y}{d x}=\cos (\mathrm{y}-\mathrm{x})+1$ is

Options

  1. A\mathrmex[sec⁡(y−x)−tan⁡(y−x)]=c\mathrme^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}
  2. B\mathrmex[sec⁡(y−x)+tan⁡(y−x)]=c\mathrme^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})+\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}
  3. C\mathrmexsec⁡(y−x)tan⁡(y−x)=c\mathrme^{\mathrm{x}} \sec (\mathrm{y}-\mathrm{x}) \tan (\mathrm{y}-\mathrm{x})=\mathrm{c}
  4. Dex=csec⁡(y−x)tan⁡(y−x)e^x=c \sec (y-x) \tan (y-x)

Correct answer

A. \mathrmex[sec⁡(y−x)−tan⁡(y−x)]=c\mathrme^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}

Step-by-step solution

dydx=cos⁡(y−x)+1\frac{d y}{d x}=\cos (y-x)+1...(1) Let y−\mathrm{y}-\mathrm{x}=\mathrm{t}⇒dydx−1=dtdx∴(1)⇒dtdx+1=cos⁡t+1\Rightarrow \frac{d y}{d x}-1=\frac{d t}{d x}\therefore(1) \Rightarrow \frac{d t}{d x}+1=\cos t+1⇒dtdx=cos⁡t⇒sec⁡t⋅dt=1⋅dx⇒∫sec⁡t⋅dt=∫1⋅dx⇒log⁡∣sec⁡t⁡+tan⁡t∣=x+c⇒−log⁡∣sec⁡t−tan⁡t∣=x+c⇒log⁡∣sec⁡t−tan⁡t∣=−x−c⇒log⁡(sec⁡(y−x)−tan⁡(y−x)=−x−c⇒sec⁡(y−x)−tan⁡(y−x)=\mathrme−x⋅\mathrme−c⇒\mathrmex(sec⁡(y−x)−tan⁡(y−x))=c\begin{aligned} & \Rightarrow \frac{d t}{d x}=\cos t \\ & \Rightarrow \sec t \cdot \mathrm{dt}=1 \cdot \mathrm{dx} \\ & \Rightarrow \int \sec t \cdot d t=\int 1 \cdot d x \\ & \Rightarrow \log |\operatorname{\sec t}+\tan t|=\mathrm{x}+\mathrm{c} \\ & \Rightarrow-\log |\sec t-\tan t|=\mathrm{x}+\mathrm{c} \\ & \Rightarrow \log |\sec t-\tan t|=-\mathrm{x}-\mathrm{c} \\ & \Rightarrow \log (\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})=-\mathrm{x}-\mathrm{c} \\ & \Rightarrow \sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})=\mathrme^{-\mathrm{x}} \cdot \mathrme^{-\mathrm{c}} \\ & \Rightarrow \mathrme^{\mathrm{x}}(\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x}))=\mathrm{c}\end{aligned}$$
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