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owen b's question at Yahoo! Answers regarding a first order homogenous ODE
- Thread starter MarkFL
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Hello owen b,
We are given to solve:
\(\displaystyle \frac{dy}{dt}=\frac{t^3}{y^3}+\frac{y}{t}\)
I would first express the ODE as:
\(\displaystyle \frac{dy}{dt}=\left(\frac{y}{t} \right)^{-3}+\frac{y}{t}\)
Now, use the substitution:
\(\displaystyle u=\frac{y}{t}\implies y=ut\implies\frac{dy}{dt}=u+\frac{du}{dt}t\)
And our ODE become:
\(\displaystyle u+\frac{du}{dt}t=u^{-3}+u\)
\(\displaystyle \frac{du}{dt}t=u^{-3}\)
Separating variables and integrating (noting $t\ne0$), we obtain:
\(\displaystyle \int u^3\,du=\int\frac{dt}{t}\)
\(\displaystyle \frac{u^4}{4}=\ln|t|+C\)
\(\displaystyle u^4=\ln\left(t^4 \right)+C\)
Back-substituting for $u$, we obtain the implicit solution:
\(\displaystyle y^4=t^4\left(\ln\left(t^4 \right)+C \right)\)
We are given to solve:
\(\displaystyle \frac{dy}{dt}=\frac{t^3}{y^3}+\frac{y}{t}\)
I would first express the ODE as:
\(\displaystyle \frac{dy}{dt}=\left(\frac{y}{t} \right)^{-3}+\frac{y}{t}\)
Now, use the substitution:
\(\displaystyle u=\frac{y}{t}\implies y=ut\implies\frac{dy}{dt}=u+\frac{du}{dt}t\)
And our ODE become:
\(\displaystyle u+\frac{du}{dt}t=u^{-3}+u\)
\(\displaystyle \frac{du}{dt}t=u^{-3}\)
Separating variables and integrating (noting $t\ne0$), we obtain:
\(\displaystyle \int u^3\,du=\int\frac{dt}{t}\)
\(\displaystyle \frac{u^4}{4}=\ln|t|+C\)
\(\displaystyle u^4=\ln\left(t^4 \right)+C\)
Back-substituting for $u$, we obtain the implicit solution:
\(\displaystyle y^4=t^4\left(\ln\left(t^4 \right)+C \right)\)