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The general solution of the differential equation (x2 + x + 1) dy+(y2 + y+ l)dx = 0 is (x+y+ 1)=A(1 + Bx + Cy+Dxy) where B, C and D are constants and A is parameter.
What is C equal to ?
- a)1
- b)-1
- c)2
- d)None o f these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The general solution of the differential equation (x2+ x + 1) dy+(y2+ ...
(x2 + x + l ) dy + (y2 + y + l ) dx = 0
(x2 + x + l)dy = -(y2 + y + l)dx
(x2 + x + l)dy = -(y2 + y + l)dx
C = - l
Most Upvoted Answer
The general solution of the differential equation (x2+ x + 1) dy+(y2+ ...
Solution:
Given differential equation: (x^2 x 1) dy = (y^2 y l)dx
To find the general solution of this differential equation, we can rewrite it as:
(x^2 x 1) dy - (y^2 y l)dx = 0
Now, let's simplify the given equation further.
(x^2 x 1) dy = (y^2 y l)dx
Dividing both sides by dx, we get:
dy/dx = (y^2 y l)/(x^2 x 1)
Now, we can rewrite the equation as:
dy/(y^2 y l) = dx/(x^2 x 1)
Separating the variables, we get:
(1/(y^2 y l)) dy = (1/(x^2 x 1)) dx
Now, let's integrate both sides of the equation.
∫(1/(y^2 y l)) dy = ∫(1/(x^2 x 1)) dx
Integrating the left side with respect to y:
∫(1/(y^2 y l)) dy = ∫(1/(x^2 x 1)) dx
Applying partial fractions to the left side, we can write it as:
1/(y^2 y l) = A/(y+1) + B/(y-1)
Multiplying both sides by (y+1)(y-1), we get:
1 = A(y-1) + B(y+1)
Simplifying the equation, we get:
1 = Ay - A + By + B
Comparing the coefficients of y on both sides, we get:
A + B = 0 ...(1)
Comparing the constant terms on both sides, we get:
-A + B = 1 ...(2)
Solving equations (1) and (2) simultaneously, we get:
A = -1/2 and B = 1/2
Substituting the values of A and B back into the partial fractions expression, we get:
1/(y^2 y l) = -1/2/(y+1) + 1/2/(y-1)
Now, let's integrate the right side of the equation with respect to x:
∫(1/(x^2 x 1)) dx = ∫(1/(x+1)) dx - ∫(1/(x-1)) dx
Integrating both sides, we get:
ln|x^2 x 1| = ln|x+1| - ln|x-1| + C
where C is the constant of integration.
Now, exponentiating both sides, we get:
|x^2 x 1| = |x+1|/|x-1| * e^C
Simplifying the equation, we get:
x^2 x 1 = (x+1)/(x-1) * e^C
Now, let's simplify the right side of the equation further.
(x+1)/(x-1) = (x-1+2)/(x-1) = 1 + 2
Given differential equation: (x^2 x 1) dy = (y^2 y l)dx
To find the general solution of this differential equation, we can rewrite it as:
(x^2 x 1) dy - (y^2 y l)dx = 0
Now, let's simplify the given equation further.
(x^2 x 1) dy = (y^2 y l)dx
Dividing both sides by dx, we get:
dy/dx = (y^2 y l)/(x^2 x 1)
Now, we can rewrite the equation as:
dy/(y^2 y l) = dx/(x^2 x 1)
Separating the variables, we get:
(1/(y^2 y l)) dy = (1/(x^2 x 1)) dx
Now, let's integrate both sides of the equation.
∫(1/(y^2 y l)) dy = ∫(1/(x^2 x 1)) dx
Integrating the left side with respect to y:
∫(1/(y^2 y l)) dy = ∫(1/(x^2 x 1)) dx
Applying partial fractions to the left side, we can write it as:
1/(y^2 y l) = A/(y+1) + B/(y-1)
Multiplying both sides by (y+1)(y-1), we get:
1 = A(y-1) + B(y+1)
Simplifying the equation, we get:
1 = Ay - A + By + B
Comparing the coefficients of y on both sides, we get:
A + B = 0 ...(1)
Comparing the constant terms on both sides, we get:
-A + B = 1 ...(2)
Solving equations (1) and (2) simultaneously, we get:
A = -1/2 and B = 1/2
Substituting the values of A and B back into the partial fractions expression, we get:
1/(y^2 y l) = -1/2/(y+1) + 1/2/(y-1)
Now, let's integrate the right side of the equation with respect to x:
∫(1/(x^2 x 1)) dx = ∫(1/(x+1)) dx - ∫(1/(x-1)) dx
Integrating both sides, we get:
ln|x^2 x 1| = ln|x+1| - ln|x-1| + C
where C is the constant of integration.
Now, exponentiating both sides, we get:
|x^2 x 1| = |x+1|/|x-1| * e^C
Simplifying the equation, we get:
x^2 x 1 = (x+1)/(x-1) * e^C
Now, let's simplify the right side of the equation further.
(x+1)/(x-1) = (x-1+2)/(x-1) = 1 + 2
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