Convolution output signal transient correction

I'm just starting out in signal analysis and I've come across this effect. When I use convolution of a sinusoid with any other, either triangular pulse, rectangular pulse or decreasing exponential (regardless of whether they are causal) a transient segment appears in the output signal. Is this effect of the convolution? How can it be solved? If I add initial conditions or conditions for t<0 will this effect be corrected?

Is this effect of the convolution?

Yes.

If you convolve two finite signals of length $$N$$ and $$K$$, the resulting signal will have a length of $$M = N + K + 1$$. Let's take a look at the convolution sum for an FIR filter

$$y[n] = \sum_{k = 0}^{K-1} h[k]\cdot x[n-k]$$

We can see that for $$n = 0$$ the sum goes back to all the way to input sample $$x[-K+1]$$. Since we have declared $$x[n]$$ to be finite on $$[0,N-1]$$ most convolution algorithms just assume zeros, i.e. $$x[n] = 0, n < 0$$. You can think about this as "pre-pending" $$K-1$$ zeros to the input signals. That creates the transient.

How can it be solved? If I add initial conditions or conditions for t<0 will this effect be corrected?

Yes. Accurate initial conditions will fix this. Along the same lines: you can just take a longer chunk of input signal and then simply discard the initial transient.

However, depending on the application, that's not always possible and in some cases you just have to live with it. There is no "one size fits all" solution.