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Use example in Differential Equation to explain the math myth (SOME PEOPLE HAVE A "MATH MIND" AND SOME DON'T.)

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Answer:

dx/dt=5x−3

Explanation:

Example 1

Solve the ordinary differential equation (ODE)

dxdt=5x−3

for x(t).

Solution: Using the shortcut method outlined in the introduction to ODEs, we multiply through by dt and divide through by 5x−3:

dx5x−3=dt.

We integrate both sides

∫dx5x−315log|5x−3|5x−3x=∫dt=t+C1=±exp(5t+5C1)=±15exp(5t+5C1)+3/5.

Letting C=15exp(5C1), we can write the solution as

x(t)=Ce5t+35.

We check to see that x(t) satisfies the ODE:

dxdt=5Ce5t5x−3=5Ce5t+3−3=5Ce5t.

Both expressions are equal, verifying our solution.

Example 2

Solve the ODE combined with initial condition:

dxdtx(2)=5x−3=1.

Solution: This is the same ODE as example 1, with solution

x(t)=Ce5t+35.

We just need to use the initial condition x(2)=1 to determine C.

C must satisfy

1=Ce5⋅2+35,

so it must be

C=25e−10.

Our solution is

x(t)=25e5(t−2)+35.

You can verify that x(2)=1.

Example 3

Solve the ODE with initial condition:

dydxy(2)=7y2x3=3.

Solution: We multiply both sides of the ODE by dx, divide both sides by y2, and integrate:

∫y−2dy−y−1y=∫7x3dx=74x4+C=−174x4+C.

The general solution is

y(x)=−174x4+C.

Verify the solution:

dydx=ddx(−174x4+C)=7x3(74x4+C)2.

Given our solution for y, we know that

y(x)2=(−174x4+C)2=1(74x4+C)2.

Therefore, we see that indeed

dydx=7x3(74x4+C)2=7x3y2.

The solution satisfies the ODE.

To determine the constant C, we plug the solution into the equation for the initial conditions y(2)=3:

3=−17424+C.

The constant C is

C=−2813=−853,

and the final solution is

y(x)=−174x4−853.

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