Question

Solve (1 + x^2) dy/dx + 2xy = cosx

Answer 1

GIVEN :–

• Diffrential equation –

$\\ \implies \bf(1 + x^2) \dfrac{dy}{dx} + 2xy = cosx$

TO FIND :–

• Solution of diffrential equation = ?

SOLUTION :–

$\\ \implies \bf(1 + x^2) \dfrac{dy}{dx} + 2xy = cosx$

$\\ \implies \bf\dfrac{dy}{dx} + \dfrac{2x}{(1 + x^2)}y = \dfrac{cosx}{(1 + x^2)}$

• Compare with –

$\\ \implies \bf\dfrac{dy}{dx} +Py = Q$

• So –

$\\ \implies \bf P =\dfrac{2x}{(1 + x^2)} \: \: and \: \: Q = \dfrac{cosx}{(1 + x^2)}$

• We know that –

$\\ \implies \bf I.F. = e^{\int P.dx}$

$\\ \implies \bf I.F. = e^{\int \frac{2x}{(1 + x^2)}dx}$

$\\ \implies \bf I.F. = e^{ \log{(1 + x^2)}}$

$\\ \implies \large{ \boxed{\bf I.F. = 1 + {x}^{2} }}$

• Solution :–

$\\ \implies \bf y(I.F.) = \int(I.F)Q.dx + c$

$\\ \implies \bf y(1 + {x}^{2} ) = \int\dfrac{cosx}{(1 + x^2)} \times (1 + {x}^{2}) .dx + c$

$\\ \implies \bf y(1 + {x}^{2} ) = \int\cos(x) .dx + c$

$\\ \implies \large { \boxed{\bf y(1 + {x}^{2} ) = \sin(x) + c}}$

Answer 2

Answer:

(1+x 2 ) dx

___ +2xy=cos x

dy

dx 2xy cos x

__ + ___ == _____

dy 1+x² 1+x²

dx

__ + Py = Q

dy

P= 2x y cos x

___ , Q == _____

1+x² 1+x²

2x

If == e∫ ____

1+x²

∴IF= 1 + x ²

y × IF = ∫Q × IF dx+c

cos x

y×(1+x ² )=∫____ ×( 1+x 2 ) dx + c

1+x ²

y×(1+x 2 )=∫cosxdx+c

y(1+x 2 )=sinx+c