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Solve the following differential equation. dydx=x2y+y

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प्रश्न

Solve the following differential equation.

`dy/dx = x^2 y + y`

बेरीज
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उत्तर

`dy/dx = x^2 y + y = (x^2 +1)y`

∴ `1/y dy = (x^2 + 1)dx`

Integrating on both sides, we get

` int 1/y dy  = int (x^2+1) dx`

∴ `log  | y | = x^3/3 + x + c`

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पाठ 8: Differential Equation and Applications - Exercise 8.3 [पृष्ठ १६५]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.3 | Q 1.1 | पृष्ठ १६५

संबंधित प्रश्‍न

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(y2 + 1) dx − (x2 + 1) dy = 0


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\[\frac{dy}{dx} + 1 = e^{x + y}\]

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\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


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\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]


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The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.


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Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


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