Help solving an ordinary differential equation
$begingroup$
I have to solve the following equation:
$$x'+frac{2}{t}x=-t^9x^5 tag{1}$$
I ave recognised that this is a Bernoulli's equation, hence I have done the substitution $z=x^{1-5}=x^{-4}$, however I reach a point where I have the following:
$$
4z'+frac{2}{t}z+t^9z^{17}=0 tag{2}
$$
where I have assumed that $x(t)>0$ and so $z(t)>0$. I'm stuck at how to solve $(2)$.
ordinary-differential-equations
asked Jan 13 at 18:25
BidonBidon
1218
$endgroup$
add a comment |
$begingroup$
I have to solve the following equation:
$$x'+frac{2}{t}x=-t^9x^5 tag{1}$$
I ave recognised that this is a Bernoulli's equation, hence I have done the substitution $z=x^{1-5}=x^{-4}$, however I reach a point where I have the following:
$$
4z'+frac{2}{t}z+t^9z^{17}=0 tag{2}
$$
where I have assumed that $x(t)>0$ and so $z(t)>0$. I'm stuck at how to solve $(2)$.
ordinary-differential-equations
asked Jan 13 at 18:25
BidonBidon
1218
$endgroup$
add a comment |
$begingroup$
I have to solve the following equation:
$$x'+frac{2}{t}x=-t^9x^5 tag{1}$$
I ave recognised that this is a Bernoulli's equation, hence I have done the substitution $z=x^{1-5}=x^{-4}$, however I reach a point where I have the following:
$$
4z'+frac{2}{t}z+t^9z^{17}=0 tag{2}
$$
where I have assumed that $x(t)>0$ and so $z(t)>0$. I'm stuck at how to solve $(2)$.
ordinary-differential-equations
asked Jan 13 at 18:25
BidonBidon
1218
$endgroup$
I have to solve the following equation:
$$x'+frac{2}{t}x=-t^9x^5 tag{1}$$
I ave recognised that this is a Bernoulli's equation, hence I have done the substitution $z=x^{1-5}=x^{-4}$, however I reach a point where I have the following:
$$
4z'+frac{2}{t}z+t^9z^{17}=0 tag{2}
$$
where I have assumed that $x(t)>0$ and so $z(t)>0$. I'm stuck at how to solve $(2)$.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 13 at 18:25
BidonBidon
1218
asked Jan 13 at 18:25
BidonBidon
1218
asked Jan 13 at 18:25
BidonBidon
1218
asked Jan 13 at 18:25
BidonBidon
1218
asked Jan 13 at 18:25
BidonBidon
1218
1218
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Write your equation in the form
$$frac{-4x'(t)}{x(t)^5}-frac{8}{tx(t)^4}=-4t^9$$ now we substitute
$$v(t)=frac{1}{x(t)^4}$$ so we get $$v'(t)-frac{8v(t)}{t}=4t^9$$ computing $$mu(t)=e^{-intfrac{8}{t}dt}=frac{1}{t^8}$$ and we get
$$frac{v'(t)}{t^8}-frac{8v(t)}{t^9}=4t$$ and we get
$$intfrac{d}{dt}left(frac{v(t)}{t^8}right)dt=int4tdt$$
Can you finish?
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
add a comment |
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1 Answer
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1 Answer
1
active
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votes
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active
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votes
$begingroup$
Write your equation in the form
$$frac{-4x'(t)}{x(t)^5}-frac{8}{tx(t)^4}=-4t^9$$ now we substitute
$$v(t)=frac{1}{x(t)^4}$$ so we get $$v'(t)-frac{8v(t)}{t}=4t^9$$ computing $$mu(t)=e^{-intfrac{8}{t}dt}=frac{1}{t^8}$$ and we get
$$frac{v'(t)}{t^8}-frac{8v(t)}{t^9}=4t$$ and we get
$$intfrac{d}{dt}left(frac{v(t)}{t^8}right)dt=int4tdt$$
Can you finish?
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
add a comment |
$begingroup$
Write your equation in the form
$$frac{-4x'(t)}{x(t)^5}-frac{8}{tx(t)^4}=-4t^9$$ now we substitute
$$v(t)=frac{1}{x(t)^4}$$ so we get $$v'(t)-frac{8v(t)}{t}=4t^9$$ computing $$mu(t)=e^{-intfrac{8}{t}dt}=frac{1}{t^8}$$ and we get
$$frac{v'(t)}{t^8}-frac{8v(t)}{t^9}=4t$$ and we get
$$intfrac{d}{dt}left(frac{v(t)}{t^8}right)dt=int4tdt$$
Can you finish?
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
add a comment |
$begingroup$
Write your equation in the form
$$frac{-4x'(t)}{x(t)^5}-frac{8}{tx(t)^4}=-4t^9$$ now we substitute
$$v(t)=frac{1}{x(t)^4}$$ so we get $$v'(t)-frac{8v(t)}{t}=4t^9$$ computing $$mu(t)=e^{-intfrac{8}{t}dt}=frac{1}{t^8}$$ and we get
$$frac{v'(t)}{t^8}-frac{8v(t)}{t^9}=4t$$ and we get
$$intfrac{d}{dt}left(frac{v(t)}{t^8}right)dt=int4tdt$$
Can you finish?
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
Write your equation in the form
$$frac{-4x'(t)}{x(t)^5}-frac{8}{tx(t)^4}=-4t^9$$ now we substitute
$$v(t)=frac{1}{x(t)^4}$$ so we get $$v'(t)-frac{8v(t)}{t}=4t^9$$ computing $$mu(t)=e^{-intfrac{8}{t}dt}=frac{1}{t^8}$$ and we get
$$frac{v'(t)}{t^8}-frac{8v(t)}{t^9}=4t$$ and we get
$$intfrac{d}{dt}left(frac{v(t)}{t^8}right)dt=int4tdt$$
Can you finish?
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
answered Jan 13 at 18:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
79.1k42867
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
add a comment |
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
$begingroup$
Yeah, getting to that substitution was the key. So simple... Thank you
$endgroup$
– Bidon
Jan 13 at 18:44
add a comment |
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