Question

Solve the following problems.

(a) \( \frac{d y}{d t}=2+\frac{y}{t} \)

(b)  \( 3 t+2 y \frac{d y}{d t}=y \)

Answer 1

Solution

Solve the following problems.

(a). \( \frac{d y}{d t}=2+\frac{y}{t} \)

Let \( u=\frac{y}{t} \Longrightarrow y=u t \Longrightarrow \frac{d y}{d t}=u+t \frac{d u}{d t} \)

Then \( \frac{d y}{d t}=2+\frac{y}{t} \Longrightarrow u+t \frac{d u}{d t}=2+u \Longrightarrow t \frac{d u}{d t}=2 \)

\( \Longrightarrow d u=\frac{2}{t} d t \Longrightarrow \int d u=\int \frac{2}{t} d t \)

\( \Longrightarrow u=2 \ln (t)+C \)

\( \Longrightarrow \frac{y}{t}=2 \ln (t)+C \)

(b). \( 3 t+2 y \frac{d y}{d t}=y \)

\( \)\( \iff 2y\frac{dy}{dt}=y-3t \)

\( \iff \frac{dy}{dt}=\frac{1}{2}-\frac{3}{2}\frac{t}{y} \)

\( y=tu\implies \frac{dy}{dt} =u+t\frac{du}{dt},\quad we set \)

\( u+t\frac{du}{dt}=\frac{1}{2}-\frac{3}{2u} \)

\( t\frac{du}{dt}=\frac{1}{2}-\frac{3}{2u}-u=\frac{u-3-2u^2}{2u} \)

\( \frac{2u}{u-3-2u^2}du=\frac{1}{t}dt \)

\( \int\frac{2u}{u-3-2u^2}du=\int\frac{1}{t}dt \)

\( \int \frac{-\frac{1}{2}(-4u+1)+\frac{1}{2}}{-2u^2+u-3}du=ln|t|$\hspace{9mm}\Big(Hint : \int \frac{dx}{x^2+a^2}=\frac{1}{a}arctan\frac{x}{a}\Big) \)

\( -\frac{1}{2}\int \frac{(-4u+1)}{-2u^2+u-3}du+\frac{1}{2}\int \frac{du}{-2\Big(u^2-\frac{1}{2}u+\frac{3}{2}\Big)}=ln|t| \)

\( -\frac{1}{2}ln\Big|-2u^2+u-3\Big|-\frac{1}{4}\int \frac{du}{\Big(u-\frac{1}{4}\Big)^2-\frac{1}{4}+\frac{3}{2}}=ln|t| \)

\( -\frac{1}{2}ln\Big|-2u^2+u-3\Big|-\frac{1}{4}\int \frac{du}{\Big(u-\frac{1}{4}\Big)^2+\frac{5}{4}}=ln|t| \)

\( -\frac{1}{2}ln\Big|-2u^2+u-3\Big|-\frac{1}{4}\times \frac{2}{\sqrt{5}}arctan\frac{2(u-\frac{1}{4})}{\sqrt{5}}=ln|t|+C \)

where \( u=\frac{y}{t} \)

 

Therefore. 

(a). \( \frac{y}{t}=2 \ln (t)+C \)

(b). \( -\frac{1}{2}ln\Big|-2u^2+u-3\Big|-\frac{1}{4}\times \frac{2}{\sqrt{5}}arctan\frac{2(u-\frac{1}{4})}{\sqrt{5}}=ln|t|+C \)