UPSC 2018 Maths Optional Paper 1 Q7c — Step-by-Step Solution

13 marks · Section B

Linear ODE with constant coefficients · ODEs · asked 4× in 13 yrs · Read the full method →

Question

Solve the initial value problem

y''-5y'+4y=e^{2t},\qquad y(0)=\frac{19}{12},\ y'(0)=\frac{8}{3}.

Technique

CF from factored auxiliary equation; PI by $D\!\to\!2$ substitution; solve the $2\times2$ linear system for $A,B$ .

Solution

Step 1 — Complementary function

Auxiliary equation $m^2-5m+4=(m-1)(m-4)=0\Rightarrow m=1,4$ :

y_c=A\,e^{t}+B\,e^{4t}.

Step 2 — Particular integral

$2$ is not a root, so substitute $D=2$ into $\dfrac{1}{D^2-5D+4}e^{2t}$ :

\phi(2)=4-10+4=-2\;\Rightarrow\;y_p=\frac{e^{2t}}{-2}=-\frac12e^{2t}.

Step 3 — General solution

y=A\,e^{t}+B\,e^{4t}-\tfrac12e^{2t},\qquad y'=A\,e^{t}+4B\,e^{4t}-e^{2t}.

Step 4 — Apply initial conditions

At $t=0$ :

y(0)=A+B-\tfrac12=\tfrac{19}{12}\;\Rightarrow\;A+B=\tfrac{19}{12}+\tfrac{6}{12}=\tfrac{25}{12},

y'(0)=A+4B-1=\tfrac{8}{3}\;\Rightarrow\;A+4B=\tfrac{8}{3}+1=\tfrac{11}{3}=\tfrac{44}{12}.

Subtract: $3B=\tfrac{44-25}{12}=\tfrac{19}{12}\Rightarrow B=\tfrac{19}{36}$ . Then $A=\tfrac{25}{12}-\tfrac{19}{36}=\tfrac{75-19}{36}=\tfrac{56}{36}=\tfrac{14}{9}$ .

Step 5 — Solution

Answer

\boxed{\;y=\frac{14}{9}e^{t}+\frac{19}{36}e^{4t}-\frac12e^{2t}.\;}