The solution of a linear differential equation dxdy - Px - Q where P and Q are functions of y- is-

The correct option is B x(I.F) = ∫ (I.F.)Qdy + C Consider the differential equation of type dxdy+P(y)x=Q(y) Then, the integrating factor for the abov ...

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Standard XII

Mathematics

Linear Differential Equations of First Order

The solution ...

Question

The solution of a linear differential equation $\frac{d x}{d y} + P x = Q$ where $P$ and $Q$ are functions of $y$ , is:

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

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

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y (I . F) = \int (I . F .) Q d y + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x (I . F) = \int (I . F .) Q d x + C

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Solution

The correct option is B $x (I . F) = \int (I . F .) Q d y + C$
Consider the differential equation of type
$\frac{d x}{d y} + P (y) x = Q (y)$
Then, the integrating factor for the above integral is
$I . F = e^{\int P (y) . d y}$
Hence multiplying the above integral by I.F
$e^{\int P (y) d y} . d x + P (y) x e^{\int P (y) d y} . d y = e^{\int P (y) d y} Q (y) . d y$
$d (x e .^{\int P (y) d y}) = e^{\int P (y) d y} Q (y) . d y$
$\int d (x e .^{\int P (y) d y}) = \int e^{\int P (y) d y} Q (y) . d y$
$x e .^{\int P (y) d y} = \int e^{\int P (y) d y} Q (y) . d y$
$x . I F = \int I F Q (y) . d y + C$

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Similar questions

The general solution of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q_{1}$ is

(a) $y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

(b) $y e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

(d) $x e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

Q. The general solution of differential equation

(x - y) d y + (x + y) d x = d x + d y

is
(where

C

is constant of integration )

A differential equation of the form

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

...(*)

where P & Q are functions of x. The number

e^{\int P d x}

when multiplied to R.H.S of (*) make it differential coefficient of a function of x & y, is called the integrating factor of the differential equation given by (*). Further the equation

\frac{d y}{d x} + P y = Q y^{n}

where P, Q are functions of x is reducible to linear form by substituting

y^{- n + 1}

as new dependent variable.

On the basis of the above information answer the following question.

The integrating factor of differential equation

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

Q. Find the singular solution of the differential equation

y = p x + p - p^{2}

. Where

p = \frac{d y}{d x}

Q. Find the general solution of the differential equation

y = p x + p^{2}

where

p = \frac{d y}{d x}

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The solution of a linear differential equation dxdy+Px=Q where P and Q are functions of y, is:

The solution of a linear differential equation $\frac{d x}{d y} + P x = Q$ where $P$ and $Q$ are functions of $y$ , is: