Differential Equations

Let y = y(x) be the solution of the differential equation dy/dx = (y+1)((y+1)e^(x²/2) - x), 0 < x < 2.1, with y(2) = 0. Then the value of dy/dx at x = 1 is equal to:

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Contributor-Level 10

The differential equation is rearranged to dt/dx - xt = -e? ²/², where t = 1/ (y+1).
This is a linear first-order differential equation. The integrating factor (I.F.) is e^ (∫-x dx) = e? ²/².
The solution is t * (I.F.) = ∫ Q (x) * (I.F.) dx + c.
t * e? ²/² = ∫ -e? ²/² * e? ²/² dx + c = ∫ -1 dx = -x + c.
Substituting t = 1/ (y+1) back: e? ²/² / (y+1) = -x + c.
Using the initial condition y (2) = 0:
e? ²/ (0+1) = -2 + c ⇒ c = e? ² + 2.
The solution is e? ²/² / (y+1) = 2 + e? ² - x.

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New answer posted

a month ago

Let Iₙ = ∫[e⁻¹ to e] x¹⁹ (log|x|)ⁿ dx, where n ∈ N. If (20)I₁₀ = αI₉ + βI₈, for natural numbers α and β then α - β equals to ______.

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Contributor-Level 10

Given the integral In = ∫(log|x|)^n / x^19 dx.
Let t = log|x|, which implies x = e^t and dx = e^t dt.

The integral becomes:
In = ∫ e^(-20t) * t^n dt

Using integration by parts, where u = t^n and dv = e^(-20t) dt:
In = [t^n * e^(-20t) / -20] - ∫ n*t^(n-1) * e^(-20t) / -20 dt
In = e^(-20) / -20 - (n / -20) * In-1
20 * In = -e^(-20) + n * In-1

For n = 10: 20 * I10 = e^20 - 10 * I9 (Note: There seems to be a sign inconsistency in the original document's derivation vs. standard integration by parts, the document states e^20 instead of -e^(-20) and proceeds with e^20).
For n = 9: 20 * I9 = e^20 - 9 * I8

From these two equations, we can express e

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New answer posted

a month ago

If x, y, z are in arithmetic progression with common difference d, x≠3d, and the determinant of the matrix [[3, 4√2, x], [4, 5√2, y], [5, k, z]] is zero, then the value of k² is:

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Contributor-Level 10

Given that x, y, z are in A.P., so 2y = x + z.
The determinant is:
| 3 4√2 x |
| 4 5√2 y | = 0
| 5 k z |

Apply the operation R? → R? + R? - 2R?:
The first row becomes:
(3 + 5 - 24) (4√2 + k - 25√2) (x + z - 2y)
= 0 (k - 6√2) (0)
So the determinant becomes:
| 0 k-6√2 0 |
| 4 5√2 y | = 0
| 5 k z |

Expanding along the first row:
-(k - 6√2)(4z - 5y) = 0.
This implies k - 6√2 = 0 or 4z - 5y = 0.
k = 6√2 or y = 4z/5.
The condition y = 4z/5 is stated as not possible.
Therefore, k = 6√2, which means k² = (6√2)² = 36 * 2 = 72.

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New answer posted

a month ago

Choose the correct statement about two circles whose equations are given below:
x² + y² - 10x - 10y + 41 = 0
x² + y² - 22x - 10y + 137 = 0

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Contributor-Level 10

Circle 1: x² + y² - 10x - 10y + 41 = 0
Center C? = (5,5).
Radius r? = √ (5² + 5² - 41) = √ (25+25-41) = √9 = 3.

Circle 2: x² + y² - 22x - 10y + 137 = 0
Center C? = (11,5).
Radius r? = √ (11² + 5² - 137) = √ (121+25-137) = √9 = 3.

Distance between centers d (C? , C? ) = √ (11-5)² + (5-5)²) = √ (6²) = 6.

Sum of radii r? + r? = 3 + 3 = 6.
Since the distance between the centers is equal to the sum of their radii, the circles touch externally at one point.

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New answer posted

a month ago

The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is:

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Contributor-Level 10

Given equations:
3x + 4y = 9
y = mx + 1 (assuming this from the substitution)

Substitute y into the first equation:
3x + 4 (mx + 1) = 9
3x + 4mx + 4 = 9
x (3 + 4m) = 5
x = 5 / (3 + 4m)

For x to be an integer, (3 + 4m) must be a divisor of 5, i.e., ±1, ±5.

3 + 4m = 1 => 4m = -2 => m = -1/2
3 + 4m = -1 => 4m = -4 => m = -1 (Integer value)
3 + 4m = 5 => 4m = 2 => m = 1/2
3 + 4m = -5 => 4m = -8 => m = -2 (Integer value)

The two integral values for m are -1 and -2.

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New answer posted

a month ago

Which of the following is true for y(x) that satisfies the differential equation dy/dx = xy - 1 + x - y; y(0) = 0:

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Contributor-Level 10

dy/dx = xy - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)
⇒ ∫ dy / (1 + y) = ∫ (x - 1) dx ⇒ ln (1 + y) = x²/2 - x + C
For y (0) = 0, c = 0. ∴ ln (1 + y) = x²/2 - x
1 + y = e^ (x²/2 - x)
y (1) = e^ (1/2 - 1) - 1 = e^ (-1/2) - 1

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New answer posted

a month ago

If y = y(x) is the solution of the differential equation (5+e?)/(2+y) dy/dx + e? = 0 satisfying y(0) = 1, then a value of y(log? 13) is:

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Contributor-Level 10

R (-1+2r, 3-2r, -r)
dr's of PR are (2 - 2r, -1+2r, -3+r)
Then 2 (2-2r) + 2 (-1+2r) + 1 (3-r) = 0
9-9r = 0 ⇒ r = 1
R (1,1, -1)
then a+1=2, b+2=2, c-3=-2
a=1, b=0, c=1
∴ a+b+c = 2

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New answer posted

a month ago

If y = (2/π x - 1) cosec x is the solution of the differential equation, dy/dx + p(x)y = 2/π cosec x, 0 < x < /2, then the function p(x) is equal to:

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Raj Pandey

Contributor-Level 9

y = (2/π x - 1)cosec x
dy/dx = (2/π)cosec x - (2/π x - 1)cosec x cot x
⇒ dy/dx + (2/π x - 1)cosec x cot x = 2/π cosec x
⇒ dy/dx + ycot x = 2/π cosec x
This is a linear differential equation. The integrating factor P (x) is the coefficient of y.
⇒ P (x) = cot x

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New answer posted

a month ago

The solution of differential equation dy/dx + (y+3x)/loge(y+3x) + 3 = 0 is:
(where C is a constant of integration.)

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Contributor-Level 10

dy/dx = (y-3x)/ln (y-3x) - 3
dy/dx + 3 = (y+3x)/ln (y+3x)
d/dx (y+3x) = (y+3x)/ln (y+3x)
∫ (ln (y+3x)/ (y+3x) d (y+3x) = ∫ dx
Let ln (y+3x) = t
1/ (y+3x) d (y+3x) = dt
∫ tdt = ∫ dx
t²/2 = x+c
(ln (y+3x)²/2 = x+c

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New answer posted

a month ago

Let y = y(x) be the solution of the differential equation, xy' − y = x²(xcos x + sin x), x > 0. If y(π) = π, then y''(π/2) + y(π/2) is equal to:

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