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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 24 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 24 Maths Textbook Solution.

Answers (1)

Answer:y=log|\tan ^{2}+2\tan x+5|+C

Hint:  Use substitution method

Given:\left(\tan ^{2} x+2 \tan x+5\right) \frac{d y}{d x}=2(1+\tan x) \sec ^{2} x

Solution: \left(\tan ^{2} x+2 \tan x+5\right) \frac{d y}{d x}=2(1+\tan x) \sec ^{2} x

              \Rightarrow \quad d y=\frac{\left.2(1+\tan x) \sec ^{2} x\right)}{\left(\tan ^{2} x+2 \tan x+5\right)} d x

Integrating both sides, we get,

\Rightarrow \quad\int d y=\frac{\left.2(1+\tan x) \sec ^{2} x\right)}{\left(\tan ^{2} x+2 \tan x+5\right)} d x

Let,\tan ^{2}x+2\tan x+5=t

Differentiate w.r.t x

\begin{aligned} &\left(2 \tan x \sec ^{2} x+2 \sec ^{2} x\right) d x=d t \\ &\Rightarrow 2(1+\tan x) \sec ^{2} x d x=d t \\ &\Rightarrow \int d y=\int \frac{1}{t} d t \\ &\Rightarrow y=\log |t|+C \\ &\Rightarrow y=\log \left|\tan ^{2}+2 \tan x+5\right|+C \end{aligned}

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