A differential equation and various initial conditions are specified. Sketch the graphs of the solutions satisfying these initia
l conditions. For each exercise, put all graphs on one pair of axes.
a. dy/dt=3y(y-2); y(0)=1; y(-2)= -1; y(0)=3; y(0)=2.
b. dy/dt=cos y; y(0)=0; y(-1)=1; y(0)= -pi/2; y(0)=pi.
c. dv/dt= -v2-2v-2; v(0)=0; v(1)=1; v(0)=1
a. dy/dt=3y(y-2); y(0)=1; y(-2)= -1; y(0)=3; y(0)=2.
b. dy/dt=cos y; y(0)=0; y(-1)=1; y(0)= -pi/2; y(0)=pi.
c. dv/dt= -v2-2v-2; v(0)=0; v(1)=1; v(0)=1
1 answer:
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Both A and C would be solutions to the equation.
In order to solve for this you must first get the equation equal to 0.
2x^2+5x+8=6 ----> subtract 6 from both sides
2x^2 + 5x + 2 = 0
Now knowing this we can use the coefficients of each one in descending order of power as a, b and c.
a = 2 (because it is the coefficient to x^2)
b = 5 (because it is the coefficient to x)
c = 2 (because it is the end number)
Now we can plug these values into the quadratic equation.
or -1/2 for the first answer
or -2 for the second answer
In order to solve for this you must first get the equation equal to 0.
2x^2+5x+8=6 ----> subtract 6 from both sides
2x^2 + 5x + 2 = 0
Now knowing this we can use the coefficients of each one in descending order of power as a, b and c.
a = 2 (because it is the coefficient to x^2)
b = 5 (because it is the coefficient to x)
c = 2 (because it is the end number)
Now we can plug these values into the quadratic equation.