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

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

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

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

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

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Solution

The correct option is D $x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$
Given DE is $\frac{d x}{d y} + P_{1} x = Q_{1}$
which is an exact DE, whose integrating factor is given by
$I F = e^{\int P_{1} d y}$
General solution of this DE is given by
$x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

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

\frac{d x}{d y} + P_{1} x = Q_{1}

y e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C

y e^{\int P_{1} d x} = \int \{Q_{1} e^{\int P_{1} d x}\} d x + C

x e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C

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

y e^{x / y} d x = (x e^{x / y} + y^{2} sin y) d y

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

y e^{x / y} d x = (x e^{x / y} + y^{2} sin y) d y

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