Question

Solve the initial value problem

9(t+1)dy/dt−7y=14t,


for t>−1t>−1 with y(0)=12. y=?

Answer 1

$9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}$

Look for an integrating factor $\mu(t)$:

$\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}$

Multiply both sides by $\mu$:

$(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}$

Condense the left side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}$

Integrate both sides:

$(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt$

For the integral on the right, substitute

$u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du$

$\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du$

$\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C$

$\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)$

$\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C$

$\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}$

Given that $y(0)=12$, we get

$12=9+C\implies C=3$

$\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}$