The correct option is A e^ y=e^x+1/3e^x^3+c. dy/dx= e^x+y+x^2e^x^3+y e^ ydy= ( e^x+x^2e^x^3 )dx Integrating, ∫ e^ ydy=∫ ( e^x+x^2e^x^3 )dx ...

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Variable Separable Method

dy/dx= ex+y+x...

Question

$\frac{d y}{d x} = e^{x + y} + x^{2} e^{x^{3} + y}$

- e^{- y} = e^{x} + \frac{1}{3} e^{x^{3}} + c .

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e^{- y} = e^{x} + \frac{1}{3} e^{x^{3}} + c .

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- e^{y} = e^{x} + \frac{1}{3} e^{x^{3}} + c .

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- e^{- y} = e^{x} - \frac{1}{3} e^{x^{3}} + c .

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Solution

The correct option is A $- e^{- y} = e^{x} + \frac{1}{3} e^{x^{3}} + c .$
$\frac{d y}{d x} = e^{x + y} + x^{2} e^{x^{3} + y}$
$e^{- y} d y = (e^{x} + x^{2} e^{x^{3}}) d x$
Integrating,
$\int e^{- y} d y = \int (e^{x} + x^{2} e^{x^{3}}) d x$
$∴ - e^{- y} = e^{x} + \int x^{2} e^{x^{3}} d x + c .$
Let $I = \int x^{2} e^{x^{3}} d x$
Put $x^{3} = t ∴ 3 x^{2} d x = d t$
$∴ I = \frac{1}{3} \int e^{t} d t = \frac{1}{3} e^{t} = \frac{1}{3} e^{^{x^{3}}}$
Thus required solution is, $- e^{- y} = e^{x} + \frac{1}{3} e^{x^{3}} + c .$

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Similar questions

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
(a) $e^{x} + e^{- y} = C$
(b) $e^{x} + e^{y} = C$
(c) $e^{- x} + e^{y} = C$
(d) $e^{- x} + e^{- y} = C$

Q. The general solution of the differential equation

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

, is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

Q. If

e^{x} + e^{y} = e^{x + y}

Show that

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

The solution of the equation $\frac{d y}{d x} - e^{x - y} + x^{2} e^{- y}$ is
[MP PET 2004]

Q. If

e^{x} (x + e) = e,

show that

\frac{d y}{d x} = e^{x} + y + 1

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dydx=ex+y+x2ex3+y

The correct option is A −e−y=ex+13ex3+c.dydx=ex+y+x2ex3+ye−ydy=(ex+x2ex3)dxIntegrating,∫e−ydy=∫(ex+x2ex3)dx∴−e−y=ex+∫x2ex3dx+c.Let I=∫x2ex3dxPut x3=t∴3x2dx=dt ∴I=13∫etdt=13et=13ex3Thus required solution is, −e−y=ex+13ex3+c.

$\frac{d y}{d x} = e^{x + y} + x^{2} e^{x^{3} + y}$