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The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 is
- a)tan-1x+log(1+x2)+tan-1y+log(1+y2)=c
- b)tan-1x-(1/2)log(1+x2)+tan-1y-(1/2)log(1+y2)=c
- c)tan-1x+(1/2)log(1+x2)+tan-1y+(1/2)log(1+y2)=c
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan-1x+...
Solution:
Given equation: (1-x^2)(1-y)dy - (1-x)(1-y^2)dx = 0
To solve this equation, we can rewrite it as:
(1-x^2)dy - (1-x)(1-y^2)dx = (1-y)dy - (1-y^2)dx
Now let's solve it step by step:
Step 1: Separate variables
(1-x^2)dy = (1-y)(dy - (1-y)dx)
Step 2: Divide both sides by (1-x^2)(1-y)
dy/dx = (1-y)/(1-x^2)
Step 3: Rewrite the equation in terms of partial fractions
(1-y)/(1-x^2) = A/(1-x) + B/(1+x)
Multiplying both sides by (1-x^2), we get:
1-y = A(1+x) + B(1-x)
Expanding, we have:
1-y = A + Ax + B - Bx
Comparing coefficients of x and constants, we get:
A - B = 0 (1)
A + B = 1 (2)
Solving equations (1) and (2), we find A = B = 1/2
Therefore, (1-y)/(1-x^2) = 1/2(1/(1-x) + 1/(1+x))
Step 4: Integrate both sides
∫(1-y)/(1-x^2) dx = ∫(1/2)(1/(1-x) + 1/(1+x)) dx
Using partial fraction decomposition, we get:
∫(1-y)/(1-x^2) dx = (1/2)∫(1/(1-x) + 1/(1+x)) dx
= (1/2)(ln|1-x| - ln|1+x|) + C1
Step 5: Integrate both sides
∫(1-y)/(1-x^2) dx = (1/2)(ln|1-x| - ln|1+x|) + C1
∫dy = (1/2)(ln|1-x| - ln|1+x|) + C1
Integrating both sides, we get:
y = (1/2)(ln|1-x| - ln|1+x|) + C2
Combining the constants of integration, we get:
y = (1/2)(ln|1-x| - ln|1+x|) + C
Therefore, the solution of the given equation is:
tan^(-1)(x) * (1/2)ln|1-x^2| + tan^(-1)(y) * (1/2)ln|1-y^2| = C
Hence, the correct answer is option C.
Given equation: (1-x^2)(1-y)dy - (1-x)(1-y^2)dx = 0
To solve this equation, we can rewrite it as:
(1-x^2)dy - (1-x)(1-y^2)dx = (1-y)dy - (1-y^2)dx
Now let's solve it step by step:
Step 1: Separate variables
(1-x^2)dy = (1-y)(dy - (1-y)dx)
Step 2: Divide both sides by (1-x^2)(1-y)
dy/dx = (1-y)/(1-x^2)
Step 3: Rewrite the equation in terms of partial fractions
(1-y)/(1-x^2) = A/(1-x) + B/(1+x)
Multiplying both sides by (1-x^2), we get:
1-y = A(1+x) + B(1-x)
Expanding, we have:
1-y = A + Ax + B - Bx
Comparing coefficients of x and constants, we get:
A - B = 0 (1)
A + B = 1 (2)
Solving equations (1) and (2), we find A = B = 1/2
Therefore, (1-y)/(1-x^2) = 1/2(1/(1-x) + 1/(1+x))
Step 4: Integrate both sides
∫(1-y)/(1-x^2) dx = ∫(1/2)(1/(1-x) + 1/(1+x)) dx
Using partial fraction decomposition, we get:
∫(1-y)/(1-x^2) dx = (1/2)∫(1/(1-x) + 1/(1+x)) dx
= (1/2)(ln|1-x| - ln|1+x|) + C1
Step 5: Integrate both sides
∫(1-y)/(1-x^2) dx = (1/2)(ln|1-x| - ln|1+x|) + C1
∫dy = (1/2)(ln|1-x| - ln|1+x|) + C1
Integrating both sides, we get:
y = (1/2)(ln|1-x| - ln|1+x|) + C2
Combining the constants of integration, we get:
y = (1/2)(ln|1-x| - ln|1+x|) + C
Therefore, the solution of the given equation is:
tan^(-1)(x) * (1/2)ln|1-x^2| + tan^(-1)(y) * (1/2)ln|1-y^2| = C
Hence, the correct answer is option C.
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