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6.5 Example

Example.

Find the unique solution to the equation d⁢xd⁢t=t⁢e2⁢t\frac{dx}{dt}=te^{2t} such that x⁢(0)=34x(0)=\frac{3}{4}.

Solution. We first find the general solution to the equation. Integrating by parts, the indefinite integral is

∫t⁢e2⁢t⁢d⁢t=12⁢t⁢e2⁢t-12⁢∫e2⁢t⁢d⁢t=12⁢t⁢e2⁢t-14⁢e2⁢t+c.\int te^{2t}\,dt={\frac{1}{2}te^{2t}-\frac{1}{2}\int e^{2t}\,dt=}{\frac{1}{2}% te^{2t}-\frac{1}{4}e^{2t}+c.}

Then to find the value of cc, we simply equate x⁢(0)=-14+c=34x(0)={-\frac{1}{4}+c}=\frac{3}{4} to obtain c=1.c={1.} Thus x⁢(t)=(12⁢t-14)⁢e2⁢t+1.x(t)={\left(\frac{1}{2}t-\frac{1}{4}\right)e^{2t}+1.}