Mister Exam
Differential equation x(y²+1)dx+y(x²+1)dy=0
The teacher will be very surprised to see your correct solution 😉
Solve as a separable equation
Solve the separable equation x (y(x)^2 + 1) + (dy(x))/(dx) (x^2 + 1) y(x) = 0:
Solve for (dy(x))/(dx):
(dy(x))/(dx) = -(x (y(x)^2 + 1))/((x^2 + 1) y(x))
Divide both sides by -(y(x)^2 + 1)/y(x):
-((dy(x))/(dx) y(x))/(y(x)^2 + 1) = x/(x^2 + 1)
Integrate both sides with respect to x:
integral-((dy(x))/(dx) y(x))/(y(x)^2 + 1) dx = integral x/(x^2 + 1) dx
Evaluate the integrals:
-1/2 log(y(x)^2 + 1) = 1/2 log(x^2 + 1) + c_1, where c_1 is an arbitrary constant.
Solve for y(x):
y(x) = -sqrt(e^(-2 c_1) - x^2 - 1)/sqrt(x^2 + 1) or y(x) = sqrt(e^(-2 c_1) - x^2 - 1)/sqrt(x^2 + 1)
Simplify the arbitrary constants:
Answer: |
| y(x) = -sqrt(-x^2 + c_1)/sqrt(x^2 + 1) or y(x) = sqrt(-x^2 + c_1)/sqrt(x^2 + 1)
Solve as a Bernoulli's equation
Solve Bernoulli's equation x (y(x)^2 + 1) + (dy(x))/(dx) (x^2 + 1) y(x) = 0:
Rewrite the equation:
(dy(x))/(dx) (x^2 + 1) y(x) + x y(x)^2 = -x
Divide both sides by 1/2 (x^2 + 1):
2 (dy(x))/(dx) y(x) + (2 x y(x)^2)/(x^2 + 1) = -(2 x)/(x^2 + 1)
Let v(x) = y(x)^2, which gives (dv(x))/(dx) = 2 y(x) (dy(x))/(dx):
(dv(x))/(dx) + (2 x v(x))/(x^2 + 1) = -(2 x)/(x^2 + 1)
Let μ(x) = e^( integral(2 x)/(x^2 + 1) dx) = x^2 + 1.
Multiply both sides by μ(x):
(x^2 + 1) (dv(x))/(dx) + (2 x) v(x) = -2 x
Substitute 2 x = d/(dx) (x^2 + 1):
(x^2 + 1) (dv(x))/(dx) + d/(dx) (x^2 + 1) v(x) = -2 x
Apply the reverse product rule f (dg)/(dx) + g (df)/(dx) = d/(dx) (f g) to the left-hand side:
d/(dx) ((x^2 + 1) v(x)) = -2 x
Integrate both sides with respect to x:
integral d/(dx) ((x^2 + 1) v(x)) dx = integral-2 x dx
Evaluate the integrals:
(x^2 + 1) v(x) = -x^2 + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = x^2 + 1:
v(x) = (-x^2 + c_1)/(x^2 + 1)
Solve for y(x) in v(x) = y(x)^2:
Answer: |
| y(x) = -sqrt(-x^2 + c_1)/sqrt(x^2 + 1) or y(x) = sqrt(-x^2 + c_1)/sqrt(x^2 + 1)
Solve as an exact equation
Solve x (y(x)^2 + 1) + (dy(x))/(dx) (x^2 + 1) y(x) = 0:
Let P(x, y) = x (y^2 + 1) and Q(x, y) = (x^2 + 1) y.
This is an exact equation, because (dP(x, y))/(dy) = 2 x y = (dQ(x, y))/(dx).
Define f(x, y) such that (df(x, y))/(dx) = P(x, y) and (df(x, y))/(dy) = Q(x, y).
Then, the solution will be given by f(x, y) = c_1, where c_1 is an arbitrary constant.
Integrate (df(x, y))/(dx) with respect to x in order to find f(x, y):
f(x, y) = integral(y^2 + 1) x dx = x^2/2 + (x^2 y^2)/2 + g(y) where g(y) is an arbitrary function of y.
Differentiate f(x, y) with respect to y in order to find g(y):
(df(x, y))/(dy) = d/(dy) (x^2/2 + (y^2 x^2)/2 + g(y)) = x^2 y + (dg(y))/(dy)
Substitute into (df(x, y))/(dy) = Q(x, y):
y x^2 + (dg(y))/(dy) = y (x^2 + 1)
Solve for (dg(y))/(dy):
(dg(y))/(dy) = y
Integrate (dg(y))/(dy) with respect to y:
g(y) = integral y dy = y^2/2
Substitute g(y) into f(x, y):
f(x, y) = x^2/2 + y^2/2 + (y^2 x^2)/2
The solution is f(x, y) = c_1:
x^2/2 + y^2/2 + (y^2 x^2)/2 = c_1
Solve for y:
y(x) = -sqrt(-x^2 + 2 c_1)/sqrt(x^2 + 1) or y(x) = sqrt(-x^2 + 2 c_1)/sqrt(x^2 + 1)
Simplify the arbitrary constants:
Answer: |
| y(x) = -sqrt(-x^2 + c_1)/sqrt(x^2 + 1) or y(x) = sqrt(-x^2 + c_1)/sqrt(x^2 + 1)
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