@ Solve x √y dx + (1+y) √(1+x) dy = 0.
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Answer:
How do I solve this differential equation (1+y) dx - (1+x) dy = 0?
This is a textbook example of a separable differential equation.
The equation given is: [math](1 + y)dx - (1 + x)dy = 0[/math]
This can also be written as: [math](1 + y)dx = (1 + x)dy[/math]
Therefore: [math]\frac{dx}{1 + x} = \frac{dy}{1 + y}[/math]
We can integrate both sides: [math]{\displaystyle \int} \frac{dx}{1 + x} = {\displaystyle \int} \frac{dy}{1 + y}[/math]
[math]ln(1 + x) + C = ln(1 + y)[/math]
Take [math]e[/math] to the power of both sides: [math]e^{ln(1 + x) + c_1} = e^{ln(1 + y)}[/math]
[math]e^{c_1}(1 + x) = 1 + y[/math]
[math]y = c_2x + (c_2 - 1), c_2 = e^{c_1}[/math]
Since we have no need for further manipulations, I will call the constant [math]c_2[/math] as [math]K[/math], so: [math]y = Kx + (K - 1)[/math], [math]K[/math] being a real constant.
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Can you find? (1+y) dx-(1-x) dy=0?
If x(1+y) ½+y(1+x) ½=0 than prove that dy/dx = —(1+x)—²?
[math] (1+y)dx-(1+x)dy=0[/math]
[math]\text{ Or } [/math]
[math] \dfrac{dx}{1+x}= \dfrac{dy}{1+y}[/math]
[math] \text{ Integrating both sides, we get}[/math]
[math]\log|1+x|=\log|1+y|+\log C \text{ where } \log C \text{ is an arbitrary constant }[/math]
[math] \text{ Raising both sides to the e-th power we get}[/math]
[math] 1+y =C(1+x)[/math]
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Separate the y's from the x' by treating this as an equality between fractions: add the dy term to both sides, then divide the two equal quantities by the common denominator of the quantities in parentheses ( always make a note that this denominator should not equal 0 ), then integrate both sides, remembering to add the integration constant-- you only need one.
The result is, ln( 1 + x ) + C = ln ( 1 + y ) . Solving for y, we get: y = e^[ ln ( 1 + x ) + C ] - 1 .
We can fiddle with the integration constant letting, C = ln k . So, y = k ( 1 + x ) - 1, within the limitations imposed by taking logs
We need to get all of the x terms with dx and the y terms with dy.
[math] (y+1) dx = (x+1) dy [/math]
[math] \frac{1}{x+1} dx = \frac{1}{y+1} [/math]
Now we integrate both sides.
[math] \int{\frac{1}{x+1}}dx = \ln{|x+1|} + C_x [/math]
We solve for y now.
[math] \ln{|y+1|} = \ln{|x+1|} + C [/math]
[math] y+1 = c(x+1) [/math]
The solution is [math] y(x) = c(x+1) - 1 [/math].
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That’s not really a differential equation, but OK. Re write it as dy/(1+y) = dx/(1+x). This integrates to ln(1+y) = ln(1+x) + c. Exponentiating gives (1+y) = C(1+x) with C to be determined by the boundary conditions
Tata Crucible Campus Quiz 2021.
A2A, thanks.
By :
[math]{dy \over 1 + y} = {dx \over 1 + x}[/math].
Now integrate both sides, with appropriate limits.
(1 + y)dx - (1 + x)dy = 0
(1 + y)dx = (1 + x)dy
dy/dx = (1 + y)/(1 + x)
How do I solve [math](x+y+1) dy/dx=1[/math]?
How do you solve (x-1) dy/dx-y=0?
How do I solve the differential equation [math]y\,dy-(1-x²-y²) x\,dx=0[/math]?
Can you find? (1+y) dx-(1-x) dy=0?
If x(1+y) ½+y(1+x) ½=0 than prove that dy/dx = —(1+x)—²?
How do you solve (1-x) dy-(3+y) dx=0?
How do I solve the first order differential equation: dy/dx= y+1/x-1,y(0) =1?
How do I solve the differential equation [math]x\sqrt{1-y^2}\,dx + y\sqrt{1-x^2}\,dy = 0[/math]?
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