solve dydx = x3 + xy, given y = 2, when x

Aditya Oke

Aditya Oke

Answered question

2022-07-15

solve dydx = x3 + xy, given y = 2, when x = 1

Answer & Explanation

star233

star233

Skilled2023-05-29Added 403 answers

To solve the differential equation dydx=x3+xy, we can use separation of variables.
First, let's rewrite the equation in a more convenient form:
dydxxy=x3
Next, we separate the variables by moving all terms involving y to one side and all terms involving x to the other side:
dydxxy=x3
dydx=x3+xy
Now, we can write the equation as:
dydx=x(x2+y)
To solve this, we can integrate both sides with respect to x:
dydxdx=(x(x2+y))dx
Integrating the left side with respect to x gives us:
y=(x(x2+y))dx
To integrate the right side, we expand the expression:
y=(x3+xy)dx
y=x3dx+xydx
y=x44+x2y2+C
Here, C is the constant of integration.
Now, we can use the initial condition given to find the value of C. We are given that when x=1, y=2. Substituting these values into the equation:
2=144+12(2)2+C
2=14+1+C
2=54+C
Simplifying, we find:
C=254=34
Now, substituting this value of C back into the equation:
y=x44+x2y2+34
Therefore, the solution to the differential equation dydx=x3+xy, with the initial condition y=2 when x=1, is:
y=x44+x2y2+34

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