Question : solve ode y”=3y’-10y y(0)=1 y'(0)=-1
Solution :
Given homogeneous linear differential equation \[y”=3y’-10y\] \[ y(0)=1,\ y'(0)=-1\] \[y”-3y’+10y=0 \] (1) So it’s auxiliary equation is \[\lambda^2-3\lambda+10=0\] \[\lambda^2-5\lambda+2\lambda+10=0\] \[\lambda\left(\lambda-5\right)+2\left(\lambda-5\right)=0\] \[\left(\lambda+2\right)\left(\lambda-5\right)=0\] It’s roots are \[\lambda_1=-2 , \lambda_2=5\] Hence the general solution of the differential equation \[y(x)=c_1e^{-2x}+c_2e^{5x} \] (2) Substituting x =0 and using first initial condition \[y\left(0\right)=1\] \[1=c_1+c_2\] (3) Differentiating eqn (2) \[y^\prime\left(x\right)=-2c_1e^{-2x}+5c_2e^{5x} \] (4) Substituting x=0 and using second initial condition \[y^\prime\left(0\right)=-1\] \[-1=-2c_1+5c_2 \] (5) And solving these equations we get \[c_1=\frac{6}{7}\ ,\ \ c_2=\frac{1}{7}\] So the particular solution of the differential equation is given as \[y(x)=\frac{6}{7}e^{-2x}+\frac{5}{6}e^{5x}\]