Mixing
Solve the differential equation $t^2y''+3ty'+y=frac{1}{t}$
$begingroup$
Solve the equation
$$t^2y''+3ty'+y=frac{1}{t}$$ with $t gt 0$
My try:
The given differential equation is actually of second order:
Let us use the substitution:
$ty=p$
$$ty'+y=frac{dp}{dt}-(1)$$
$$ty''+2y'=frac{d^2p}{dt^2}$$
$$t^2y''+2ty'=tfrac{d^2p}{dt^2}-(2)$$
Adding (1) and (2) we get:
$$t^2y''+3ty'+y=tfrac{d^2p}{dt^2}+frac{dp}{dt}$$
Hence the equation is now:
$$tfrac{d^2p}{dt^2}+frac{dp}{dt}=frac{1}{t}$$
Let us use another substitution:
$frac{dp}{dt}=q$
Then the equation is:
$$tfrac{dq}{dt}+q=frac{1}{t}$$
$$frac{dq}{dt}+frac{q}{t}=frac{1}{t^2}$$
Which is a Linear first order differential equation with integrating factor given by:
$$I(t)=e^{int frac{dt}{t}}=e^{ln|t|}=|t|=t$$
The solution is:
$$qt=int frac{dt}{t}+C$$
$$qt=ln t+C$$
$$q=frac{ln t+C}{t}$$
Now we get:
$$frac{dp}{dt}=frac{ln t+C}{t}$$
Integrating we get:
$$p=int frac{ln t}{t}+int frac{C}{t}+D$$
$$p=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
Hence the final solution is:
$$ty=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
$$y=frac{frac{1}{2}left(ln tright)^2+Cln(t)+D}{t}$$
Where $C$ and $D$ are constants:
Is there any different approach?
integration algebra-precalculus ordinary-differential-equations derivatives
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
$endgroup$
add a comment |
$begingroup$
Solve the equation
$$t^2y''+3ty'+y=frac{1}{t}$$ with $t gt 0$
My try:
The given differential equation is actually of second order:
Let us use the substitution:
$ty=p$
$$ty'+y=frac{dp}{dt}-(1)$$
$$ty''+2y'=frac{d^2p}{dt^2}$$
$$t^2y''+2ty'=tfrac{d^2p}{dt^2}-(2)$$
Adding (1) and (2) we get:
$$t^2y''+3ty'+y=tfrac{d^2p}{dt^2}+frac{dp}{dt}$$
Hence the equation is now:
$$tfrac{d^2p}{dt^2}+frac{dp}{dt}=frac{1}{t}$$
Let us use another substitution:
$frac{dp}{dt}=q$
Then the equation is:
$$tfrac{dq}{dt}+q=frac{1}{t}$$
$$frac{dq}{dt}+frac{q}{t}=frac{1}{t^2}$$
Which is a Linear first order differential equation with integrating factor given by:
$$I(t)=e^{int frac{dt}{t}}=e^{ln|t|}=|t|=t$$
The solution is:
$$qt=int frac{dt}{t}+C$$
$$qt=ln t+C$$
$$q=frac{ln t+C}{t}$$
Now we get:
$$frac{dp}{dt}=frac{ln t+C}{t}$$
Integrating we get:
$$p=int frac{ln t}{t}+int frac{C}{t}+D$$
$$p=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
Hence the final solution is:
$$ty=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
$$y=frac{frac{1}{2}left(ln tright)^2+Cln(t)+D}{t}$$
Where $C$ and $D$ are constants:
Is there any different approach?
integration algebra-precalculus ordinary-differential-equations derivatives
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
$endgroup$
add a comment |
$begingroup$
Solve the equation
$$t^2y''+3ty'+y=frac{1}{t}$$ with $t gt 0$
My try:
The given differential equation is actually of second order:
Let us use the substitution:
$ty=p$
$$ty'+y=frac{dp}{dt}-(1)$$
$$ty''+2y'=frac{d^2p}{dt^2}$$
$$t^2y''+2ty'=tfrac{d^2p}{dt^2}-(2)$$
Adding (1) and (2) we get:
$$t^2y''+3ty'+y=tfrac{d^2p}{dt^2}+frac{dp}{dt}$$
Hence the equation is now:
$$tfrac{d^2p}{dt^2}+frac{dp}{dt}=frac{1}{t}$$
Let us use another substitution:
$frac{dp}{dt}=q$
Then the equation is:
$$tfrac{dq}{dt}+q=frac{1}{t}$$
$$frac{dq}{dt}+frac{q}{t}=frac{1}{t^2}$$
Which is a Linear first order differential equation with integrating factor given by:
$$I(t)=e^{int frac{dt}{t}}=e^{ln|t|}=|t|=t$$
The solution is:
$$qt=int frac{dt}{t}+C$$
$$qt=ln t+C$$
$$q=frac{ln t+C}{t}$$
Now we get:
$$frac{dp}{dt}=frac{ln t+C}{t}$$
Integrating we get:
$$p=int frac{ln t}{t}+int frac{C}{t}+D$$
$$p=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
Hence the final solution is:
$$ty=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
$$y=frac{frac{1}{2}left(ln tright)^2+Cln(t)+D}{t}$$
Where $C$ and $D$ are constants:
Is there any different approach?
integration algebra-precalculus ordinary-differential-equations derivatives
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
$endgroup$
Solve the equation
$$t^2y''+3ty'+y=frac{1}{t}$$ with $t gt 0$
My try:
The given differential equation is actually of second order:
Let us use the substitution:
$ty=p$
$$ty'+y=frac{dp}{dt}-(1)$$
$$ty''+2y'=frac{d^2p}{dt^2}$$
$$t^2y''+2ty'=tfrac{d^2p}{dt^2}-(2)$$
Adding (1) and (2) we get:
$$t^2y''+3ty'+y=tfrac{d^2p}{dt^2}+frac{dp}{dt}$$
Hence the equation is now:
$$tfrac{d^2p}{dt^2}+frac{dp}{dt}=frac{1}{t}$$
Let us use another substitution:
$frac{dp}{dt}=q$
Then the equation is:
$$tfrac{dq}{dt}+q=frac{1}{t}$$
$$frac{dq}{dt}+frac{q}{t}=frac{1}{t^2}$$
Which is a Linear first order differential equation with integrating factor given by:
$$I(t)=e^{int frac{dt}{t}}=e^{ln|t|}=|t|=t$$
The solution is:
$$qt=int frac{dt}{t}+C$$
$$qt=ln t+C$$
$$q=frac{ln t+C}{t}$$
Now we get:
$$frac{dp}{dt}=frac{ln t+C}{t}$$
Integrating we get:
$$p=int frac{ln t}{t}+int frac{C}{t}+D$$
$$p=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
Hence the final solution is:
$$ty=frac{1}{2}left(ln tright)^2+Cln(t)+D$$
$$y=frac{frac{1}{2}left(ln tright)^2+Cln(t)+D}{t}$$
Where $C$ and $D$ are constants:
Is there any different approach?
integration algebra-precalculus ordinary-differential-equations derivatives
integration algebra-precalculus ordinary-differential-equations derivatives
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
asked Jan 16 at 14:23
Umesh shankarUmesh shankar
2,72231220
2,72231220
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
The given ODE reduces to the following solution
begin{align*}
& t^{2}y^{primeprime} + 3ty^{prime} + y = frac{1}{t} Leftrightarrow (t^{2}y^{primeprime} + 2ty^{prime}) + (ty^{prime} + y) = frac{1}{t} Leftrightarrow (t^{2}y^{prime})^{prime} + (ty)^{prime} = frac{1}{t} Leftrightarrow\
&int (t^{2}y^{prime})^{prime}mathrm{d}t + int (ty)^{prime}mathrm{d}t = intfrac{1}{t}mathrm{d}t Leftrightarrow t^{2}y^{prime} + ty = ln|t| + c Leftrightarrow ty^{prime} + y = frac{ln|t| + c}{t} Leftrightarrow\
& (ty)^{prime} = frac{ln|t| + c}{t} Leftrightarrow int(ty)^{prime}mathrm{d}t = intleft(frac{ln|t|}{t} + frac{c}{t}right)mathrm{d}t Leftrightarrow y(t) = frac{ln^{2}|t|}{2t} + frac{cln|t|}{t} + frac{k}{t}
end{align*}
answered Jan 16 at 15:03
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to
$$
0=r(r-1)+3r+1=(r+1)^2.
$$
The basis solutions for this double root are thus $t^{-1}$, $t^{-1}ln t$.
As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(ln t)^2$. Insertion into the equation allows then to compute $C$.
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The given ODE reduces to the following solution
begin{align*}
& t^{2}y^{primeprime} + 3ty^{prime} + y = frac{1}{t} Leftrightarrow (t^{2}y^{primeprime} + 2ty^{prime}) + (ty^{prime} + y) = frac{1}{t} Leftrightarrow (t^{2}y^{prime})^{prime} + (ty)^{prime} = frac{1}{t} Leftrightarrow\
&int (t^{2}y^{prime})^{prime}mathrm{d}t + int (ty)^{prime}mathrm{d}t = intfrac{1}{t}mathrm{d}t Leftrightarrow t^{2}y^{prime} + ty = ln|t| + c Leftrightarrow ty^{prime} + y = frac{ln|t| + c}{t} Leftrightarrow\
& (ty)^{prime} = frac{ln|t| + c}{t} Leftrightarrow int(ty)^{prime}mathrm{d}t = intleft(frac{ln|t|}{t} + frac{c}{t}right)mathrm{d}t Leftrightarrow y(t) = frac{ln^{2}|t|}{2t} + frac{cln|t|}{t} + frac{k}{t}
end{align*}
answered Jan 16 at 15:03
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
The given ODE reduces to the following solution
begin{align*}
& t^{2}y^{primeprime} + 3ty^{prime} + y = frac{1}{t} Leftrightarrow (t^{2}y^{primeprime} + 2ty^{prime}) + (ty^{prime} + y) = frac{1}{t} Leftrightarrow (t^{2}y^{prime})^{prime} + (ty)^{prime} = frac{1}{t} Leftrightarrow\
&int (t^{2}y^{prime})^{prime}mathrm{d}t + int (ty)^{prime}mathrm{d}t = intfrac{1}{t}mathrm{d}t Leftrightarrow t^{2}y^{prime} + ty = ln|t| + c Leftrightarrow ty^{prime} + y = frac{ln|t| + c}{t} Leftrightarrow\
& (ty)^{prime} = frac{ln|t| + c}{t} Leftrightarrow int(ty)^{prime}mathrm{d}t = intleft(frac{ln|t|}{t} + frac{c}{t}right)mathrm{d}t Leftrightarrow y(t) = frac{ln^{2}|t|}{2t} + frac{cln|t|}{t} + frac{k}{t}
end{align*}
answered Jan 16 at 15:03
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
The given ODE reduces to the following solution
begin{align*}
& t^{2}y^{primeprime} + 3ty^{prime} + y = frac{1}{t} Leftrightarrow (t^{2}y^{primeprime} + 2ty^{prime}) + (ty^{prime} + y) = frac{1}{t} Leftrightarrow (t^{2}y^{prime})^{prime} + (ty)^{prime} = frac{1}{t} Leftrightarrow\
&int (t^{2}y^{prime})^{prime}mathrm{d}t + int (ty)^{prime}mathrm{d}t = intfrac{1}{t}mathrm{d}t Leftrightarrow t^{2}y^{prime} + ty = ln|t| + c Leftrightarrow ty^{prime} + y = frac{ln|t| + c}{t} Leftrightarrow\
& (ty)^{prime} = frac{ln|t| + c}{t} Leftrightarrow int(ty)^{prime}mathrm{d}t = intleft(frac{ln|t|}{t} + frac{c}{t}right)mathrm{d}t Leftrightarrow y(t) = frac{ln^{2}|t|}{2t} + frac{cln|t|}{t} + frac{k}{t}
end{align*}
answered Jan 16 at 15:03
user1337user1337
46110
$endgroup$
The given ODE reduces to the following solution
begin{align*}
& t^{2}y^{primeprime} + 3ty^{prime} + y = frac{1}{t} Leftrightarrow (t^{2}y^{primeprime} + 2ty^{prime}) + (ty^{prime} + y) = frac{1}{t} Leftrightarrow (t^{2}y^{prime})^{prime} + (ty)^{prime} = frac{1}{t} Leftrightarrow\
&int (t^{2}y^{prime})^{prime}mathrm{d}t + int (ty)^{prime}mathrm{d}t = intfrac{1}{t}mathrm{d}t Leftrightarrow t^{2}y^{prime} + ty = ln|t| + c Leftrightarrow ty^{prime} + y = frac{ln|t| + c}{t} Leftrightarrow\
& (ty)^{prime} = frac{ln|t| + c}{t} Leftrightarrow int(ty)^{prime}mathrm{d}t = intleft(frac{ln|t|}{t} + frac{c}{t}right)mathrm{d}t Leftrightarrow y(t) = frac{ln^{2}|t|}{2t} + frac{cln|t|}{t} + frac{k}{t}
end{align*}
answered Jan 16 at 15:03
user1337user1337
46110
edited Jan 16 at 17:08
answered Jan 16 at 15:03
user1337user1337
46110
answered Jan 16 at 15:03
user1337user1337
46110
answered Jan 16 at 15:03
user1337user1337
46110
46110
add a comment |
add a comment |
$begingroup$
The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to
$$
0=r(r-1)+3r+1=(r+1)^2.
$$
The basis solutions for this double root are thus $t^{-1}$, $t^{-1}ln t$.
As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(ln t)^2$. Insertion into the equation allows then to compute $C$.
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
$endgroup$
add a comment |
$begingroup$
The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to
$$
0=r(r-1)+3r+1=(r+1)^2.
$$
The basis solutions for this double root are thus $t^{-1}$, $t^{-1}ln t$.
As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(ln t)^2$. Insertion into the equation allows then to compute $C$.
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
$endgroup$
add a comment |
$begingroup$
The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to
$$
0=r(r-1)+3r+1=(r+1)^2.
$$
The basis solutions for this double root are thus $t^{-1}$, $t^{-1}ln t$.
As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(ln t)^2$. Insertion into the equation allows then to compute $C$.
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
$endgroup$
The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to
$$
0=r(r-1)+3r+1=(r+1)^2.
$$
The basis solutions for this double root are thus $t^{-1}$, $t^{-1}ln t$.
As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(ln t)^2$. Insertion into the equation allows then to compute $C$.
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
answered Jan 16 at 15:12
LutzLLutzL
58.8k42056
58.8k42056
add a comment |
add a comment |
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