(2y-1)dy/dx=(3x^2+1) given that x=1 when y=2
plz show step thanks
2)
(a)find the general solution of the equation (x-2)dy/dx+3y(x-1)/(x+1)=1
b)given the boundary condition y=5 when x=-1,find the particular solution of the condition given in (a)
a little help would do thanks i have the basic of calculus
see your earlier post for #1
For #2, review your techniques for 1st order linear ODEs, and see whether you can get the solution given here:
http://www.wolframalpha.com/input/?i=(x-2)+dy%2Fdx+%3D+1-3y(x-1)%2F(x%2B1)
Solve the homogeneous equation using partial fractions, and then work into the general solution.
To solve the given differential equation (2y-1)dy/dx = (3x^2+1), we'll follow these steps:
Step 1: Separate the variables.
Rearrange the equation to separate the variables, placing the dy terms on one side and the dx terms on the other side:
(2y - 1)dy = (3x^2 + 1)dx
Step 2: Integrate both sides.
Integrate both sides of the equation with respect to their respective variables:
∫(2y - 1)dy = ∫(3x^2 + 1)dx
Step 3: Perform the integrations.
∫(2y - 1)dy can be split into two separate integrals:
∫2y dy - ∫dy
∫(3x^2 + 1)dx can be integrated as:
∫3x^2 dx + ∫1 dx
Step 4: Evaluate the integrals.
For ∫2y dy:
Using the power rule of integration, the integral becomes:
y^2 + C1
For ∫3x^2 dx:
Using the power rule of integration, the integral becomes:
x^3 + C2
For ∫1 dx:
The integral of 1 with respect to x is simply x.
Remember to add the constant of integration, which is represented by the C1 and C2 terms.
Step 5: Rewrite the equation using the integrated results.
The equation now becomes:
y^2 - y = x^3 + x + C
Step 6: Apply the initial condition.
Given that x = 1 when y = 2, substitute these values into the equation and solve for the constant of integration C:
2^2 - 2 = 1^3 + 1 + C
4 - 2 = 2 + C
2 = 2 + C
C = 0
Step 7: Write the final solution.
Substitute the value of C back into the equation:
y^2 - y = x^3 + x + 0
Simplifying, we get:
y^2 - y = x^3 + x
This is the general solution of the given differential equation.