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If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation is
- a)2x3 + 2xy + y2 = c
- b)2x3 + 6xy + 3y2 = c
- c)2x + 2xy + y2 = c
- d)x2 + 2xy + y2 = c
Correct answer is option 'B'. Can you explain this answer?
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If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if...
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If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if...
Given: g(x, y)dx + (x - y)dy = 0
To determine if the given differential equation is exact, we need to check if the partial derivatives of the function g(x, y) satisfy the equality:
∂g/∂y = ∂(x - y)/∂x
Differentiating the expression on both sides with respect to x, we get:
∂²g/∂x∂y = ∂²(x - y)/∂x²
Since the second-order mixed partial derivatives are equal, the differential equation is exact.
To find the general solution, we need to find a function f(x, y) such that the total differential of f(x, y) is equal to the given differential equation.
We can rewrite the given equation as:
g(x, y)dx + xdy - ydy = 0
Simplifying further:
g(x, y)dx + xdy = ydy
Comparing this with the total differential form:
df = Mdx + Ndy
df = g(x, y)dx + xdy
We can equate the coefficients of dx and dy:
M = g(x, y)
N = x
Integrating M with respect to x gives us f(x, y):
f(x, y) = ∫g(x, y)dx
We can differentiate f(x, y) with respect to y and equate it to N:
∂f/∂y = ∂(∫g(x, y)dx)/∂y = x
Taking the partial derivative of f(x, y) with respect to y, we can find f(x, y):
∂f/∂y = x
Integrating x with respect to y, we get:
f(x, y) = xy + h(x)
where h(x) is an arbitrary function of x.
Now, we need to find g(x, y) using the initial condition g(x, 0) = x².
Substituting y = 0 in f(x, y), we get:
f(x, 0) = x*0 + h(x) = h(x)
Comparing this with g(x, y), we have:
g(x, y) = h(x)
Therefore, g(x, y) = x².
Finally, the general solution of the given differential equation is:
xy + h(x) = c
or
2x³ - 6xy + 3y² = c
Hence, the correct answer is option B: 2x³ - 6xy + 3y² = c.
To determine if the given differential equation is exact, we need to check if the partial derivatives of the function g(x, y) satisfy the equality:
∂g/∂y = ∂(x - y)/∂x
Differentiating the expression on both sides with respect to x, we get:
∂²g/∂x∂y = ∂²(x - y)/∂x²
Since the second-order mixed partial derivatives are equal, the differential equation is exact.
To find the general solution, we need to find a function f(x, y) such that the total differential of f(x, y) is equal to the given differential equation.
We can rewrite the given equation as:
g(x, y)dx + xdy - ydy = 0
Simplifying further:
g(x, y)dx + xdy = ydy
Comparing this with the total differential form:
df = Mdx + Ndy
df = g(x, y)dx + xdy
We can equate the coefficients of dx and dy:
M = g(x, y)
N = x
Integrating M with respect to x gives us f(x, y):
f(x, y) = ∫g(x, y)dx
We can differentiate f(x, y) with respect to y and equate it to N:
∂f/∂y = ∂(∫g(x, y)dx)/∂y = x
Taking the partial derivative of f(x, y) with respect to y, we can find f(x, y):
∂f/∂y = x
Integrating x with respect to y, we get:
f(x, y) = xy + h(x)
where h(x) is an arbitrary function of x.
Now, we need to find g(x, y) using the initial condition g(x, 0) = x².
Substituting y = 0 in f(x, y), we get:
f(x, 0) = x*0 + h(x) = h(x)
Comparing this with g(x, y), we have:
g(x, y) = h(x)
Therefore, g(x, y) = x².
Finally, the general solution of the given differential equation is:
xy + h(x) = c
or
2x³ - 6xy + 3y² = c
Hence, the correct answer is option B: 2x³ - 6xy + 3y² = c.
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If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation isa)2x3 + 2xy + y2 = cb)2x3 + 6xy + 3y2 = cc)2x + 2xy + y2 = cd)x2 + 2xy + y2 = cCorrect answer is option 'B'. Can you explain this answer?
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