Answer to Question #293872 in Differential Equations for Narmi

Question

Show that y=c1epower x +c2e power 2x is the general solution of y"-3y'+2y=0 on any interval and find the particular solution for which y(0) =1

Answer 1

We want to show that $\displaystyle y=c_1e^x+c_2e^{2x} \text{ is the general solution of }y\prime\prime-3y\prime+2y=0\\$ on any interval and find the particular solution for which $\displaystyle y(0)=1$.

Now,

$y=c_1e^x+c_2e^{2x}\\ y\prime=c_1e^x+2c_2e^{2x}\\ y\prime\prime=c_1+4c_2e^{2x}$

Substituting $\displaystyle y, y\prime, y\prime\prime$ into the given DE yields;

$\displaystyle (c_1e^x+4c_2e^{2x})-3(c_1e^x+2c_2e^{2x})+2(c_1e^x+c_2e^{2x})=0\\ \Rightarrow c_1e^x+4c_2e^{2x}-3c_1e^x-6c_2e^{2x}+2c_1e^x+2c_2e^{2x}=0\\ \Rightarrow e^x(c_1-3c_1+2c_1)+e^{2x}(4c_2-6c_2+2c_2)=0\\ \Rightarrow e^x(0)+e^{2x}(0)=0\\ \Rightarrow 0=0$

Hence, we have shown that $\displaystyle y=c_1e^x+c_2e^{2x}$ is the general solution of the given DE on any interval.

Next, at $\displaystyle y(0)=1$, we have;

$\displaystyle 1=c_1+c_2\Rightarrow c_1=1-c_2$

Thus, the particular solution is for which $\displaystyle y(0)=1$ is; $\displaystyle y=c_1e^x+c_2e^{2x}=(1-c_2)e^x+c_2e^{2x}$