Initial Value Problem
dy/dx = y^2 -5y+6
y(0)=1
y' = y^2-5y+6
dy/((y-3)(y-2)) = dx
dy (1/(y-3) - 1/(y-2)) = dx
ln[(y-3)/y-2)] = x + c
(y-3)/(y-2) = c e^x
y(0) = 1, so
(1-3)/(1-2) = c
c = 2
(y-3)/(y-2) = 2e^x
Now you can rearrange things and find y as a function of x:
y = (3-4e^x)/(1-2e^x)
To solve the initial value problem dy/dx = y^2 - 5y + 6, y(0) = 1, we can use the method of separable variables.
Step 1: Separate the variables by rewriting the equation as:
dy / (y^2 - 5y + 6) = dx
Step 2: Factor the denominator:
dy / ((y - 2)(y - 3)) = dx
Step 3: Expand the left-hand side using partial fractions:
(A / (y - 2)) + (B / (y - 3)) = dy / ((y - 2)(y - 3))
(A(y - 3) + B(y - 2)) = dy
Step 4: Equate the coefficients of y on both sides:
A + B = 1
Step 5: Equate the constant terms on both sides:
-3A - 2B = 0
Solving these equations gives A = 1/2 and B = 1/2.
Step 6: Substitute the partial fractions back into the equation from step 3:
(1/2) * (1 / (y - 2)) + (1/2) * (1 / (y - 3)) = dy
Step 7: Integrate both sides of the equation:
∫((1/2) * (1 / (y - 2)) + (1/2) * (1 / (y - 3))) dy = ∫ dx
(1/2) * ln |y - 2| + (1/2) * ln |y - 3| = x + C
Step 8: Combine the logarithmic terms:
ln |(y - 2)(y - 3)| = 2x + 2C
Step 9: Remove the natural logarithm:
|(y - 2)(y - 3)| = e^(2x + 2C)
Step 10: Remove the absolute value:
(y - 2)(y - 3) = ± e^(2x + 2C)
Step 11: Solve for y:
Two cases need to be considered to account for the ± sign:
Case 1: (y - 2)(y - 3) = e^(2x + 2C)
Expanding the equation gives:
y^2 - 5y + 6 = e^(2x + 2C)
Rearranging the equation gives:
y^2 - 5y + 6 - e^(2x + 2C) = 0
Using the initial condition y(0) = 1, we substitute these values into the equation to find the value of C.
Case 2: (y - 2)(y - 3) = - e^(2x + 2C)
Expanding the equation gives:
y^2 - 5y + 6 = - e^(2x + 2C)
Rearranging the equation gives:
y^2 - 5y + 6 + e^(2x + 2C) = 0
Using the initial condition y(0) = 1, we substitute these values into the equation to find the value of C.
Upon solving for C in both cases, the general solution for y(x) can be obtained by substituting the value of C into the respective case equations.
To solve the initial value problem dy/dx = y^2 - 5y + 6 with the initial condition y(0) = 1, we can use the method of separable variables. Here's how you can go about it:
Step 1: Rewrite the differential equation in the standard form: dy/dx = y^2 - 5y + 6.
Step 2: Since we have dy/dx on one side and y-related terms on the other side, let's bring all the y-related terms to the left side and keep the dy/dx on the right side.
This gives us: dy/(y^2 - 5y + 6) = dx.
Step 3: Now we separate the variables by multiplying both sides by dx and separating the denominators.
This gives us: 1/(y^2 - 5y + 6) dy = dx.
Step 4: We can now integrate both sides of the equation.
Integrating the left side: ∫(1/(y^2 - 5y + 6)) dy.
To determine the integral of the left side, we need to factorize the denominator.
The factorization is: y^2 - 5y + 6 = (y - 2)(y - 3).
So, the integral becomes: ∫(1/((y - 2)(y - 3))) dy.
Using partial fraction decomposition, we can write 1/((y - 2)(y - 3)) as A/(y - 2) + B/(y - 3), where A and B are constants.
By multiplying through by (y - 2)(y - 3), we get: 1 = A(y - 3) + B(y - 2).
Comparing the coefficients of y, we have: 0y = -3A - 2B.
This yields: -3A - 2B = 0.
Comparing the constant terms, we have: 1 = -3A + 2B.
This yields: -3A + 2B = 1.
Solving these two equations simultaneously, we find: A = 1/5 and B = 2/5.
Therefore, 1/((y - 2)(y - 3)) = (1/5)/(y - 2) + (2/5)/(y - 3).
Now, we can integrate both sides of the equation.
Integrating: ∫(1/((y - 2)(y - 3))) dy = ∫((1/5)/(y - 2)) dy + ∫((2/5)/(y - 3)) dy.
Integrating each term, we get:
ln|y - 2| - ln|y - 3| = (1/5)ln|y - 2| + (2/5)ln|y - 3| + C,
where C is the constant of integration.
Step 5: Now, we simplify the equation and solve for y.
ln|y - 2| - ln|y - 3| = (1/5)ln|y - 2| + (2/5)ln|y - 3| + C.
Combining the ln terms, we have:
ln|y - 2| - (1/5)ln|y - 2| - (2/5)ln|y - 3| = ln|y - 3| + C.
Simplifying the equation further, we get:
(4/5)ln|y - 2| - (2/5)ln|y - 3| = ln|y - 3| + C.
Multiplying through by 5, we have:
4ln|y - 2| - 2ln|y - 3| = 5ln|y - 3| + 5C.
Combining the ln terms, we get:
ln|y - 2|^4 - ln|y - 3|^2 = ln|y - 3|^5 + ln|e^(5C)|.
Using the properties of logarithms, we can simplify this equation further:
ln|((y - 2)^4/(y - 3)^2)| = ln|((y - 3)^5)| + ln|e^(5C)|.
Since the natural logarithm of a quantity is equal to the natural logarithm of its exponential form, we can rewrite the equation as:
((y - 2)^4/(y - 3)^2) = (e^(5C))(y - 3)^5.
Simplifying further, we have:
(y - 2)^4 = (e^(5C))(y - 3)^3.
Step 6: Now that we have the general solution, we can apply the initial condition y(0) = 1 to find the specific solution.
Substituting y = 1 and x = 0 into the equation:
(1 - 2)^4 = (e^(5C))(1 - 3)^3.
Simplifying, we get:
(-1)^4 = (e^(5C))(-2)^3.
Which becomes:
1 = 8(e^(5C)).
Dividing both sides by 8, we get:
(e^(5C)) = 1/8.
Taking the natural logarithm of both sides, we have:
5C = ln(1) - ln(8),
5C = 0 - ln(8),
5C = -ln(8),
C = (-ln(8))/5.
Therefore, the specific solution to the initial value problem dy/dx = y^2 - 5y + 6, y(0) = 1 is:
(y - 2)^4 = (1/8)(y - 3)^3.