Differential Equations
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If x³dy + xydx = x²dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to:
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If x³dy + xydx = x²dy + 2ydx; y(2) = e and x > 1, then y(4) is equal to:
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x³dy + xydx = 2ydx + x²dy
=> (x³-x²)dy = (2-x)ydx
=> dy/y = (2-x)/ (x² (x-1) dx
=> ∫ (dy/y) = ∫ (2-x)/ (x² (x-1)dx
Let (2-x)/ (x² (x-1) = A/x + B/x² + C/ (x-1)
=> 2-x = A (x-1) + B (x-1) + Cx²
=> C=1, B=-2 and A=-1
=> ∫ (dy/y) = ∫ (-1/x - 2/x² + 1/ (x-1)dx
=> lny = -lnx + 2/x + ln|x-1| + C
∴ y (2) = e
=> 1 =
Upvote 
The solution curve of the differential equation, (1 + e⁻ˣ)(1 + y²)dy/dx = y², which passes through the point (0,1), is
The solution curve of the differential equation, (1 + e⁻ˣ)(1 + y²)dy/dx = y², which passes through the point (0,1), is
(1 + e? ) (1 + y²) dy/dx = y²
⇒ (1 + y? ²)dy = ( e? / (1 + e? ) ) dx
⇒ (y - 1/y) = ln (1 + e? ) + c
∴ It passes through (0,1) ⇒ c = -ln2
⇒ y² = 1 + yln ( (1+e? )/2 )
Upvote 
Let y = y(x) be the solution of the differential equation, ((2+sinx)/(y+1))dy/dx = -cosx, y>0, y(0)=1. If y(π)=a and dy/dx at x=π is b, then the ordered pair (a,b) is equal to
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Let y = y(x) be the solution of the differential equation, ((2+sinx)/(y+1))dy/dx = -cosx, y>0, y(0)=1. If y(π)=a and dy/dx at x=π is b, then the ordered pair (a,b) is equal to
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(2+sin x)/ (y+1) dy/dx = -cosx, y>0
⇒ dy/ (y+1) = -cosx/ (2+sinx) dx
By integrating both sides:
ln|y+1| = -ln|2+sinx|+lnK
⇒ y+1 = K/ (2+sinx) (y+1>0)
⇒ y (x) = K/ (2+sinx) - 1
Given y (0)=1 ⇒ 1=K/2-1 ⇒ K=4
So, y (x)=4/ (2+sinx)-1
a=y (π)=1
b=dy/dx|x=π = -cosx/ (y (x)+1)|x=π = 1
So, (a, b)= (1,1)
Upvote 
Let y = y(x) be solution of the following differential equation e?(dy/dx) - 2e?sinx + sinxcos²x = 0, y(π/2) = 0. If y(0) = log?(α + βe?²), then 4(α+β) is equal to…….
Let y = y(x) be solution of the following differential equation e?(dy/dx) - 2e?sinx + sinxcos²x = 0, y(π/2) = 0. If y(0) = log?(α + βe?²), then 4(α+β) is equal to…….
e? (dy/dx) - 2e? sinx + sinxcos²x = 0
d/dx (e? ) - (2sinx)e? = -sinxcos²x
I.F. = e^ (-∫2sinxdx) = e²cosx
Solution: e? e²cosx = -∫e²cosx sinx cos²x dx
Let cosx=t, -sinxdx=dt
∫e²? t²dt = e²? t²/2 - ∫2te²? /2 dt = e²? t²/2 - [te²? /2 - ∫e²? /2 dt] = e²? t²/2 - te²? /2 + e²? /4 + C
e^ (y+2cosx) = e²cosxcos²x/
Upvote The general solution of the differential equation √(1+x²+y²+x²y²) + xy(dy/dx)=0 is: (where C is a constant of integration)
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The general solution of the differential equation √(1+x²+y²+x²y²) + xy(dy/dx)=0 is: (where C is a constant of integration)
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√ (1+x²) (1+y²) + xy (dy/dx)=0.
√ (1+x²)/x dx + √ (1+y²)/y dy = 0.
√ (1+x²)+½ln| (√ (1+x²)-1)/ (√ (1+x²)+1)|+√ (1+y²)=C.
Upvote