Differential Equations
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If dy/dx * 2ʸ - (2ʸ-1)/(2ˣ-1) = 0, x,y > 0, y(1)=1, then y(2) is equal to:
If dy/dx * 2ʸ - (2ʸ-1)/(2ˣ-1) = 0, x,y > 0, y(1)=1, then y(2) is equal to:
dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3
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If ∫ (x²+1)eˣ / (x+1)² dx = f(x) eˣ + C, where C is a constant, then d³f/dx³ at x=1 is equal
If ∫ (x²+1)eˣ / (x+1)² dx = f(x) eˣ + C, where C is a constant, then d³f/dx³ at x=1 is equal
I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4
Upvote Let the curve y = y(x) be the solution of the differential equation, dy/dx = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is 4√8/3, then the value of y(1) is equal to.
Let the curve y = y(x) be the solution of the differential equation, dy/dx = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is 4√8/3, then the value of y(1) is equal to.
dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 =
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Let xᵏ + yᵏ = aᵏ, (a, k > 0) and dy/dx + (y/x)¹/³ = 0, then k is:
Let xᵏ + yᵏ = aᵏ, (a, k > 0) and dy/dx + (y/x)¹/³ = 0, then k is:
Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.
Upvote If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :
If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :
dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t -
Upvote Let C₁ be the curve obtained by the solution of differential equation 2xy dy/dx = y² - x², x > 0. Let the curve C₂ be the solution of 2xy/(x²-y²) = dy/dx. If both the curves pass through (1, 1), then the area enclosed by the curves C₁ and C₂ is equal to :
Let C₁ be the curve obtained by the solution of differential equation 2xy dy/dx = y² - x², x > 0. Let the curve C₂ be the solution of 2xy/(x²-y²) = dy/dx. If both the curves pass through (1, 1), then the area enclosed by the curves C₁ and C₂ is equal to :
If y = y(x) is the solution of the differential equation dy/dx + (tanx)y = sinx, 0 ≤ x ≤ π/3, with y(0) = 0, then y(π/4) equal to:
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If y = y(x) is the solution of the differential equation dy/dx + (tanx)y = sinx, 0 ≤ x ≤ π/3, with y(0) = 0, then y(π/4) equal to:
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dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C =>
Upvote 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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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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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
Upvote 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 ______.
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 ______.
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)
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