find the solution of the initial value problem y'=xye^x; y(1)=1
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To find the solution of the initial value problem y' = xy*e^x, y(1) = 1, we will use the method of separation of variables.
Step 1: Separate the variables.
Start by writing the differential equation in the form:
dy/dx = xy*e^x
Now, divide both sides by y*e^x to separate the variables:
dy/(y*e^x) = x*dx
Step 2: Integrate both sides.
Integrate the left side with respect to y and the right side with respect to x:
∫(1/y) dy = ∫x dx
The integral of 1/y with respect to y is ln|y|, and the integral of x with respect to x is (1/2)x^2:
ln|y| = (1/2)x^2 + C
where C is the constant of integration.
Step 3: Solve for y.
To find the particular solution, we need to solve for y. Exponentiate both sides with the base e:
|y| = e^((1/2)x^2 + C)
Since y(1) = 1, we can substitute this value into the equation to determine the sign of the absolute value:
|1| = e^(1/2 + C)
Taking the absolute value of 1 on the left side of the equation, we have:
1 = e^(1/2 + C)
Step 4: Solve for C.
To find the value of C, take the natural logarithm of both sides:
ln(1) = ln(e^(1/2 + C))
Since ln(1) equals 0, the equation simplifies to:
0 = (1/2) + C
Solve for C:
C = -(1/2)
Step 5: Rewrite the solution.
Replace C in the original equation with -(1/2):
|y| = e^((1/2)x^2 - (1/2))
Finally, we can write the solution as:
y = ±e^((1/2)x^2 - (1/2))
The ± indicates that there are two solutions, one with a positive y value and one with a negative y value.
Therefore, the solution to the initial value problem y' = xy*e^x, y(1) = 1 is:
y = ±e^((1/2)x^2 - (1/2))