1)Find the particular solution of the differential equation
xy'+3y=80xlnx (x>0)
satisfying the initial condition y(1)=4
2)Find the particular solution of the differential equation
y'−1y=−5x+3
satisfying the initial condition y(0)=3.
To find the particular solution of a differential equation, we will follow the general process of separation of variables, integrating both sides, and applying initial conditions.
1) Find the particular solution of the differential equation:
xy' + 3y = 80xlnx (x > 0)
satisfying the initial condition y(1) = 4.
Step 1: Separate the variables.
Rearrange the equation to isolate the terms with y and y':
xy' = -3y + 80xlnx.
Step 2: Integrate both sides.
Integrate both sides with respect to x:
∫xy' dx = ∫(-3y + 80xlnx) dx.
Using the formula for integration by parts on the left side of the equation:
∫xy' dx = xy - ∫y dx.
Apply the integration by parts on the right side:
xy - ∫y dx = ∫(-3y + 80xlnx) dx.
Step 3: Simplify and solve for y.
Rearrange the equation to solve for y:
∫(-3y + 80xlnx) dx = ∫y dx.
∫(-3y) dx + ∫(80xlnx) dx = ∫y dx.
Assuming that the integral of y will yield a C constant:
-3∫y dx + ∫(80xlnx) dx = C.
Integrate each term separately:
-3∫y dx = -3xy + c1.
∫(80xlnx) dx = c2.
Substitute the results back into the equation:
-3xy + c1 + c2 = C.
Step 4: Apply the initial condition.
Substitute x = 1 and y = 4 into the equation:
-3(1)(4) + c1 + c2 = C.
Simplify and solve for C:
-12 + c1 + c2 = C.
Since we are given the initial condition y(1) = 4, substitute the values and solve for C:
-12 + c1 + c2 = 4.
c1 + c2 = 16.
Now, we have the equation in general form for the particular solution:
-3xy + c1 + c2 = C.
By rearranging and substituting the value of c1 + c2:
-3xy = C - 16.
The particular solution is given by:
y = (C - 16) / (-3x).
2) Find the particular solution of the differential equation:
y' - y = -5x + 3
satisfying the initial condition y(0) = 3.
Step 1: Rearrange the equation
y' - y = -5x + 3
Step 2: Solve the first order linear ODE
Multiply through by the integrating factor, which is e^(-x):
e^(-x)y' - e^(-x)y = -5xe^(-x) + 3e^(-x)
Now, use the product rule on the left side:
(e^(-x)y)' = -5xe^(-x) + 3e^(-x)
Integrate both sides with respect to x:
∫(e^(-x)y)' dx = ∫(-5xe^(-x) + 3e^(-x)) dx
Using the fundamental theorem of calculus:
e^(-x)y = -5∫(xe^(-x)) dx + 3∫(e^(-x)) dx
Simplify and integrate each term:
e^(-x)y = -5∫(xe^(-x)) dx + 3e^(-x) + C
Step 3: Solve for y
Divide both sides by e^(-x):
y = -5∫(xe^(-x)) dx + 3 + Ce^(x)
Step 4: Apply the initial condition
Substitute x = 0 and y = 3 into the equation:
3 = -5∫(0e^(-0)) dx + 3 + Ce^(0)
Simplify and solve for C:
3 = -5∫(0) dx + 3 + C
Since the integral evaluates to 0, we have:
C = 3
Now, substitute the value of C back into the equation:
y = -5∫(xe^(-x)) dx + 3e^(x) + 3
The particular solution is given by:
y = -5∫(xe^(-x)) dx + 3e^(x) + 3.