3.8 Differential equations

For dy/dx=f(x)g(y), separate into dy/g(y)=f(x)dx and integrate. A linear equation dy/dx+P(x)y=Q(x) uses an integrating factor e^{∫Pdx}.

State where division is legal and retain any constant solutions lost by dividing by g(y). Differentiate the result to verify the form.

dy/dx=xy separates as dy/y=x dx, giving ln|y|=x²/2+C and y=Ae^{x²/2}; y=0 is also included by A=0.

Dividing by y can hide the zero solution, and an integrating factor is not needed for every separable equation.

Integrating an ODE introduces arbitrary constants. The order of the equation indicates how many independent conditions are normally needed to select one solution.

Keep the general solution visible, use conditions at the specified point, and check the resulting function in both the differential equation and the conditions.

Solving y′=2x gives y=x²+C; y(0)=3 selects y=x²+3.

A general solution is not one trajectory, and a condition on y′ cannot replace a missing condition on y when only one constant exists.

An initial condition such as y(x₀)=y₀ is substituted into the integrated family to determine its constant. For a second-order equation, conditions on y and y′ may be needed.

Apply the condition after integrating, not before; if the problem gives a physical initial velocity, translate it into y′(x₀) explicitly.

If y′=3x²−2 and y(1)=4, then y=x³−2x+C and C=5.

The initial point is not automatically x=0; use the coordinate stated in the question.

A solution of an ODE is a function whose derivatives satisfy the equation on the stated interval. Solve using the appropriate method, then apply initial or boundary information.

Check differentiability and any division restrictions. Substitution into the original equation is the strongest final check, especially after squaring or taking logarithms.

For y′=y, y=Ce^x satisfies the equation; y(0)=2 selects C=2.

A formula that matches one initial point is not enough; it must also satisfy the differential relation throughout the domain.