Question

solve (1+t²)y' +4ty=(1+t²)^-2; y(0)=1

Answer 1

$\text{ Solve the linear equation } \frac{d y(t)}{d t}\left(t^{2}+1\right)+4 t y(t)=\frac{1}{\left(t^{2}+1\right)^{2}}, \text{ such that } y(0)=1 :\\[4mm] \text{ Divide both sides by } t^{2}+1 :\\[1.5mm] \frac{d y(t)}{d t}+\frac{4 t y(t)}{t^{2}+1}=\frac{1}{\left(t^{2}+1\right)^{3}}\\[2mm] \text{ Let } \mu(t)=e^{\int(4 t) /\left(t^{2}+1\right) d t}=\left(t^{2}+1\right)^{2}.\\[4mm] \text{ Multiply both sides by } \mu(t) :\\[1.5mm] \left(t^{2}+1\right)^{2} \frac{d y(t)}{d t}+\left(4 t\left(t^{2}+1\right)\right) y(t)=\frac{1}{t^{2}+1}\\[2mm] \text{ Substitute } 4 t\left(t^{2}+1\right)=\frac{d}{d t}\left(\left(t^{2}+1\right)^{2}\right) :\\[2mm] \left(t^{2}+1\right)^{2} \frac{d y(t)}{d t}+\frac{d}{d t}\left(\left(t^{2}+1\right)^{2}\right) y(t)=\frac{1}{t^{2}+1}\\[4mm] \text{ Apply the reverse product rule } f \frac{d g}{d t}+g \frac{d f}{d t}=\frac{d}{d t}(f g) \text{ to the left-hand side:}\\[2mm] \frac{d}{d t}\left(\left(t^{2}+1\right)^{2} y(t)\right)=\frac{1}{t^{2}+1}$

$\text{Integrate both sides with respect to } t :\\[1.5mm] \int \frac{d}{d t}\left(\left(t^{2}+1\right)^{2} y(t)\right) d t=\int \frac{1}{t^{2}+1} d t\\[4mm] \text{Evaluate the integrals:}\\ \left(t^{2}+1\right)^{2} y(t)=\tan ^{-1}(t)+c_{1}, \text{ where } c_{1} \text{ is an arbitrary constant.}\\[4mm] \text{Divide both sides by } \mu(t)=\left(t^{2}+1\right)^{2} :\\[1.5mm] y(t)=\frac{\tan ^{-1}(t)+c_{1}}{\left(t^{2}+1\right)^{2}}\\[4mm] \text{ Solve for } c_{1} \text{ using the initial conditions:}\\[2mm] \text{ Substitute } y(0)=1 \text{ into } y(t)=\frac{\tan ^{-1}(t)+c 1}{\left(t^{2}+1\right)^{2}} :\\ c_{1}=1\\[2mm] \text{ Substitute } c_{1}=1 \text{ into } y(t)=\frac{\tan ^{-1}(t)+c_{1}}{\left(t^{2}+1\right)^{2}}\\[2mm] \text{ Answer:}\\[2mm] y(t)=\frac{\tan ^{-1}(t)+1}{\left(t^{2}+1\right)^{2}}$