If (t + 1)dx = (2x + (t + 1)3)dt and x(0) = 2 then x(1) is equal to

(t + 1)dx = (2x + (t + 1)3)dt d x d t − 2 x t + 1 = ( t + 1 ) 2 I.F. = e ∫ − 2 t + 1 d t = 1 ( t + 1 ) 2 Solution is x ( t + 1 ) 2 = ∫ 1 d t <!- [if gte mso 9]><xml> </xml><! [endif]->x = (t + c) (t + 1)2 ? x (0) = 2 then c = 2x = (t + 2) (t + 1)2 x (1) = 12

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First Order Differential Equation Differential Equations Maths Class 12th

If (t + 1)dx = (2x + (t + 1)³)dt and x(0) = 2 then x(1) is equal to

Option 1 -

Option 2 -

Option 3 -

Option 4 -

3 Views | Posted 4 weeks ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 weeks ago

Correct Option - 2

Detailed Solution:

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{d x}{d t} - \frac{2 x}{t + 1} = {(t + 1)}^{2}$

I.F. $= e^{\int - \frac{2}{t + 1} d t} = \frac{1}{{(t + 1)}^{2}}$

Solution is

$\frac{x}{{(t + 1)}^{2}} = \int 1 d t$

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

x (1) = 12

(t + 1)dx = (2x + (t + 1)3)dt <math> <mrow> <mfrac> <mrow> <mi>d</mi> <mi>x</mi> </mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> <mi>x</mi> </mrow> <mrow> <mi>t</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>t</mi> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> I.F. <math> <mrow> <mo>=</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mi>t</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mstyle> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>t</mi> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>  Solution is <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>t</mi> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mn>1</mn> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mstyle> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1820326447"> </o:OLEObject></xml><! [endif]->x = (t + c) (t + 1)2 <math> <mrow> <mo>? </mo> </mrow> </math> x (0) = 2 then c = 2x = (t + 2) (t + 1)2 x (1) = 12

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If (t + 1)dx = (2x + (t + 1)³)dt and x(0) = 2 then x(1) is equal to

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