Let y = y(x) be the solution curve of the differential equation <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>+</mo><mrow><mo>(</mo><mrow><mfrac><mrow><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>1</mn><mn>1</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>3</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mn>6</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>1</mn><mn>1</mn><mi>x</mi><mo>+</mo><mn>6</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mi>y</mi><mo>=</mo><mfrac><mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>+</mo><mn>3</mn></mrow><mo>)</mo></mrow></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>,</mo><mi>x</mi><mo>></mo><mo>−</mo><mn>1</mn><mo>,</mo></mrow></math> which passes through the point (0, 1). Then y(1) is equal to

dydx+(2x2+11x+13x3+6x2+11x+6)y=x+3x+1,x>−1IF = e∫pdx=(x+1)2(x+2)x+3∫Pdx=∫2x2+11x+13x3+6x2+11x+6dn=∫(2x+1+1x+2−1x+3)dx=ln((x+1)2⋅(x+2)/(x+3))2x2+11x+13(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+32x2+11x+13=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)x = -1⇒ 4 = 2A ⇒ A = 2x = -2⇒ -1 = -B Þ B = 1x2 – 3 ⇒ -2 = 2cc = -1y⋅(x+1)2(x+2)x+3=∫x+3x+1⋅(x+1)2(x+2)x+3dx= ∫(x+1)(x+2)dx= x33+3x22+2x+c(0,1)⇒1⋅23=cx = 1 y⋅(3)=13+32+2+23=32+3=92y = 32

Let y = y(x) be the solution curve of the differential equation&nbsp;<math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>+</mo><mrow><mo>(</mo><mrow><mfrac><mrow><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>1</mn><mn>1</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>3</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mn>6</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>1</mn><mn>1</mn><mi>x</mi><mo>+</mo><mn>6</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mi>y</mi><mo>=</mo><mfrac><mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>+</mo><mn>3</mn></mrow><mo>)</mo></mrow></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>,</mo><mi>x</mi><mo>&gt;</mo><mo>−</mo><mn>1</mn><mo>,</mo></mrow></math>&nbsp;which passes through the point (0, 1). Then y(1) is equal to

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Maths Class 12th Differential Equations

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to

Option 1 -

$\frac{1}{2}$

Option 2 -

$\frac{3}{2}$

Option 3 -

$\frac{5}{2}$

Option 4 -

$\frac{7}{2}$

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$

IF = $e^{\int p d x} = \frac{{(x + 1)}^{2} (x + 2)}{x + 3}$

$\int P d x = \int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d n = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$

$= l n ({(x + 1)}^{2} \cdot (x + 2) / (x + 3))$

$\frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

$2 x^{2} + 11 x + 13 = A (x + 2) (x + 3) + B (x + 1) (x + 3) + C (x + 1) (x + 2)$

x = -1

⇒ 4 = 2A ⇒ A = 2

x = -2

⇒ -1 = -B Þ B = 1

x₂ – 3 ⇒ -2 = 2c

c = -1

$y \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} = \int \frac{x + 3}{x + 1} \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} d x$

= $\int (x + 1) (x + 2) d x$

= $\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c$

$(0, 1) \Rightarrow 1 \cdot \frac{2}{3} = c$

x = 1 $y \cdot (3) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = \frac{3}{2} + 3 = \frac{9}{2}$

y = $\frac{3}{2}$

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Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to

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Let y = y(x) be the solution curve of the differential equation dydx+(2x2+11x+13x3+6x2+11x+6)y=(x+3)x+1,x>−1, which passes through the point (0, 1). Then y(1) is equal to

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Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to