# How do you solve #y'+3y=0# given y(0)=4?

This is a first order linear ordinary differential equation and so we can derive an Integrating Factor using:

So multiplying the DE by the IF gives:

Thanks to the product rule this can now be written as the derivative of a single product:

Which can now easily be solved to give;

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We have the separable differential equation:

Which can be rearranged as:

Integrate both sides:

Then:

Rewriting using exponent rules:

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To solve the differential equation ( y' + 3y = 0 ) with the initial condition ( y(0) = 4 ), you can use the method of separation of variables.

- Rewrite the equation in the form ( y' = -3y ).
- Separate variables by dividing both sides by ( y ) and multiplying both sides by ( dx ): ( \frac{dy}{y} = -3dx ).
- Integrate both sides:
- Integrate ( \frac{dy}{y} ) to get ( \ln|y| ).
- Integrate ( -3dx ) to get ( -3x + C ), where ( C ) is the constant of integration.

- Combine the results: ( \ln|y| = -3x + C ).
- Solve for ( y ) by exponentiating both sides: ( |y| = e^{-3x + C} ).
- Apply the initial condition ( y(0) = 4 ):
- When ( x = 0 ), ( |y| = e^C ).
- Since ( |y| = 4 ), ( e^C = 4 ).
- Therefore, ( C = \ln(4) ).

- Substitute ( C = \ln(4) ) back into the equation: ( |y| = e^{-3x + \ln(4)} ).
- Simplify and solve for ( y ):
- ( |y| = e^{\ln(4) - 3x} = 4e^{-3x} ).
- ( y = \pm 4e^{-3x} ).

- Apply the initial condition to find the particular solution:
- ( y(0) = 4 ) implies ( y = 4e^0 = 4 ).

- Hence, the solution to the initial value problem is ( y = 4e^{-3x} ).

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