-x-y-dx- where dy-dx-0 is given by

The correct option is D (x+y)|x+y|/2+c We know that, d|x|= |x| x dx Now, I=∫ |x+y|dx =∫ |x+y|( x+y ) #160; ( x+y ) #160; dx I=∫ |x+y| ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

∫|x+y|dx, whe...

Question

$\int | x + y | d x$ , where $\frac{d y}{d x} = 0$ is given by

0

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\frac{(x + y)^{2}}{2} + c

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- \frac{(x + y)^{2}}{2} + c

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\frac{(x + y) | x + y |}{2} + c

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Solution

The correct option is D $\frac{(x + y) | x + y |}{2} + c$
We know that,
$d | x | = \frac{| x |}{x} d x$
Now,
$I = \int | x + y | d x = \int | x + y | \frac{(x + y)}{(x + y)} d x I = \int \frac{| x + y |}{(x + y)} \cdot (x + y) d x L e t x + y = t ∴ d x = d t a s \frac{d y}{d x} = 0 I = \int \frac{| t |}{t} t d t$
Using integration by parts,
$I = t \int \frac{| t |}{t} d t - \int \frac{d (t)}{d t} (\int \frac{| t |}{t} t d t) d t I = t | t | - \int | t | d t + K = t | t | - I + K 2 I = t | t | + K I = \frac{t | t |}{2} + C I = \frac{(x + y) | x + y |}{2} + C$

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Similar questions

Q. If

\sqrt{y + x} + \sqrt{y - x} = c

, show that

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

Q. The differential equation whose general solution is given by,

y = (c_{1} c o s (x + c_{2})) - (c_{3} e^{(- x + c_{4})}) + (c_{5} s i n x)

where

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Which of the following differential equations has y=x as one of its particular solution?
(a) $\frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x$
(b) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x$
(c) $\frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0$
(d) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0$

Q. In the following, show that the given differential equation is homogeneous and solve each of them.
1.

(x^{2} + x y) d y = (x^{2} + y^{2}) d x

y^{'} = \frac{x + y}{x}

(x y) d y (x + y) d x = 0

(x^{2} + y^{2}) d x + 2 x y d y = 0

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

x d y . y d x = \sqrt{x^{2} + y^{2}} d x

{x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y d x = {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} x d y

x \frac{d y}{d x} - y + x sin (\frac{y}{x}) = 0

y d x + x log (\frac{y}{x}) d y - 2 x d y = 0

Q. Solve :

\frac{d y}{d x} = \frac{y}{x} + \frac{\sqrt{x^{2} + y^{2}}}{x}, x > 0.

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∫|x+y|dx, where dydx=0 is given by

The correct option is D (x+y)|x+y|2+cWe know that,d|x|=|x|xdxNow,I=∫|x+y|dx=∫|x+y|(x+y)(x+y)dxI=∫|x+y|(x+y)⋅(x+y)dxLetx+y=t∴dx=dtasdydx=0I=∫|t|ttdtUsing integration by parts,I=t∫|t|tdt−∫d(t)dt(∫|t|ttdt)dtI=t|t|−∫|t|dt+K=t|t|−I+K2I=t|t|+KI=t|t|2+CI=(x+y)|x+y|2+C

$\int | x + y | d x$ , where $\frac{d y}{d x} = 0$ is given by