MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Derivative from First Principle

Solve 1+x2d...

Question

Solve $(1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1)$

Open in App

Solution

We have,

$(1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1)$

$\Rightarrow \frac{d y}{d x} - \frac{2 x y}{(1 + x^{2})} = \frac{(x^{2} + 2) (x^{2} + 1)}{(1 + x^{2})}$

$\Rightarrow \frac{d y}{d x} - \frac{2 x y}{(1 + x^{2})} = (x^{2} + 2)$

On comparing that,

$\frac{d y}{d x} + P y = Q$

Then,

$P = - \frac{2 x}{1 + x^{2}}, Q = x^{2} + 2$

Now,

$I . F . = e^{\int p d x} = e^{\int \frac{- 2 x}{1 + x^{2}} d x} = e^{- log (1 + x^{2})} = \frac{1}{1 + x^{2}}$

So,

$y . I . F . = \int Q I . F . d x + C$

$\Rightarrow y . \frac{1}{1 + x^{2}} = \int \frac{x^{2} + 2}{1 + x^{2}} d x + C$

$\Rightarrow \frac{y}{1 + x^{2}} = \int \frac{x^{2} + 1}{1 + x^{2}} d x + \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow \frac{y}{1 + x^{2}} = x + {tan}^{- 1} x + C$

Hence, this is the answer.

Suggest Corrections

thumbs-up

0

Similar questions

(1 - x^{2}) \frac{d y}{d x} + 2 x y = x \sqrt{1 - x^{2}}

(1 - x^{2}) \frac{d y}{d x} + 2 x y = x {(1 - x^{2})}^{1 / 2}

Q. Solve the differential equation:

(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}

.

Q. Solve the following differential equation:

x^{2} \frac{d y}{d x} = y^{2} + 2 x y

y = 1

x = 1

Q. Solve the differential equation:

(1 - x^{2}) \frac{d y}{d x} - x y = \frac{1}{\sqrt{(1 - x^{2})}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Derivative of Simple Functions

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Derivative from First Principle

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon