Differential Equations
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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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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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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
Upvote 
Choose the correct statement about two circles whose equations are given below:
x² + y² - 10x - 10y + 41 = 0
x² + y² - 22x - 10y + 137 = 0
Choose the correct statement about two circles whose equations are given below:
x² + y² - 10x - 10y + 41 = 0
x² + y² - 22x - 10y + 137 = 0
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
Upvote 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:
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:
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 =
Upvote 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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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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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
Upvote 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:
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:
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
Upvote 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 < π/2, then the function p(x) is equal to:
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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 < π/2, then the function p(x) is equal to:
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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
Upvote 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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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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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
Upvote 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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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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x dy/dx - y = x² (xcosx + sinx), x > 0
dy/dx - y/x = x (xcosx+sinx) ⇒ dy/dx + Py = Q
So, I.F. = e^ (∫-1/x dx) = 1/|x| = 1/x (x > 0)
Thus, y/x = ∫ 1/x (x (xcosx+sinx)dx
⇒ y/x = xsinx + C
? y (π) = π ⇒ C = 1
So, y = x²sinx + x ⇒ (y)? /? = π²/4 + π/2
Also, dy/dx = x²cosx + 2xsinx + 1
⇒ d²y/dx² = -x²sinx + 4xco
Upvote