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# Solution - Solve the differential equation (1 + x2) dy/dx+y=e^(tan^(−1))x. - ISC (Arts) Class 12 - Mathematics

ConceptSolutions of Linear Differential Equation

#### Question

Solve the differential equation  (1 + x2) dy/dx+y=e^(tan^(−1))x.

#### Solution

(1 + x2) dy/dx+y=e^(tan^(−1))x

=>dy/dx+y/(1+x^2)=e^(tan^(−1)x)/(1+x^2)

Comparing it with the standard equation, f'(x)+yP=Q

Now, integrating factor, I.F. = e^(∫Pdx)=e^(int1/(1+x^2)dx)=e^(tan^(-1)x)

ye^(∫P(x)dx)=∫Q(x)e^(∫P(x)dx)dx+C

∴ ye^(tan^(−1) x)=∫(e^(tan^(−1) x))/(1+x^2)e^(tan^(−1) x)dx+C            .....(1)

Let I=∫(e^(tan^(−1) x))/(1+x^2)e^(tan^(−1) x)dx

Putting e^(tan^(−1) x)=t

∫1/(1+x^2)e^(tan^(−1) x)dx=dt

∴ I=∫t dt

I=t^2/2

⇒I=((e^(tan^(−1) x))^2)/2

Considering (1), we get:

ye^(tan^(−1) x)=(e^(tan^(−1) x)/2)^2+C

Is there an error in this question or solution?

#### Reference Material

Solution for question: Solve the differential equation (1 + x2) dy/dx+y=e^(tan^(−1))x. concept: Solutions of Linear Differential Equation. For the courses ISC (Arts), ISC (Science), ISC (Commerce), CBSE (Commerce), CBSE (Science), CBSE (Arts), PUC Karnataka Science
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