Needing help solving ( 1 + x^3)dy/dx + 5(x^2)y = e^x
- Thread starter mrgthiru
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This is a particular case of general equation:
y' + y*p(x) = g(x)
you need to multiply both sides by an integarting factor - e^[Q(x)] - where,
Q(x) = [integral]P(x) dx
Then th left-hand-side will become a "perfect integral".
Upon integrating both sides
y * e^[Q(x)] = [integral]{g * e^[Q(x)]} dx
It should be straight-forward.
y' + y*p(x) = g(x)
you need to multiply both sides by an integarting factor - e^[Q(x)] - where,
Q(x) = [integral]P(x) dx
Then th left-hand-side will become a "perfect integral".
Upon integrating both sides
y * e^[Q(x)] = [integral]{g * e^[Q(x)]} dx
It should be straight-forward.
Re: Could anyone help me in solving this Differential Equ
Hello, mrgthiru!
Subhotosh is partly correct . . . An integrating factor is required.
. . But the integration will be difficult.
Divide by \(\displaystyle (1\,+\,x^3):\;\;\L\frac{dy}{dx} \,+\,\frac{5x^2}{1\,+\,x^3}\cdot y \:=\:e^x\)
\(\displaystyle \L I \;=\;e^{\int\frac{5x^2}{1+x^3}dx} \;=\;e^{\frac{5}{3}\ln(1+x^3)} \;=\;e^{\ln(1+x^3)^{\frac{5}{3}}}\;=\;(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{Multiply through by }I:\;\;\L(1\,+\,x^3)^{\frac{5}{3}}\cdot\frac{dy}{dx}\,+\,\frac{5x^2}{1\,+\,x^3}\cdot(1\,+\,x^3)^{\frac{5}{3}}\cdot y \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
. . . \(\displaystyle \L(1\,+\,x^3)^{\frac{5}{3}}\cdot\frac{dy}{dx}\:+\:5x^2(1\,+\,x^3)^{\frac{2}{3}}y \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{We have: }\L\:\frac{d}{dx}\left[(1\,+\,x^3)^{\frac{5}{3}}y\right] \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{Integrate: }\L\
1\,+\,x^3)^{\frac{5}{3}}y \;=\;\underbrace{\int e^x(1\,+\,x^3)^{\frac{5}{3}}\,dx}_{\text{Good luck!}}\)
Welcome aboard, Subhotosh! . . . Good to see you here!
Hello, mrgthiru!
Subhotosh is partly correct . . . An integrating factor is required.
. . But the integration will be difficult.
Solve: \(\displaystyle \
Divide by \(\displaystyle (1\,+\,x^3):\;\;\L\frac{dy}{dx} \,+\,\frac{5x^2}{1\,+\,x^3}\cdot y \:=\:e^x\)
\(\displaystyle \L I \;=\;e^{\int\frac{5x^2}{1+x^3}dx} \;=\;e^{\frac{5}{3}\ln(1+x^3)} \;=\;e^{\ln(1+x^3)^{\frac{5}{3}}}\;=\;(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{Multiply through by }I:\;\;\L(1\,+\,x^3)^{\frac{5}{3}}\cdot\frac{dy}{dx}\,+\,\frac{5x^2}{1\,+\,x^3}\cdot(1\,+\,x^3)^{\frac{5}{3}}\cdot y \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
. . . \(\displaystyle \L(1\,+\,x^3)^{\frac{5}{3}}\cdot\frac{dy}{dx}\:+\:5x^2(1\,+\,x^3)^{\frac{2}{3}}y \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{We have: }\L\:\frac{d}{dx}\left[(1\,+\,x^3)^{\frac{5}{3}}y\right] \;=\;e^x(1\,+\,x^3)^{\frac{5}{3}}\)
\(\displaystyle \text{Integrate: }\L\
1\,+\,x^3)^{\frac{5}{3}}y \;=\;\underbrace{\int e^x(1\,+\,x^3)^{\frac{5}{3}}\,dx}_{\text{Good luck!}}\)Welcome aboard, Subhotosh! . . . Good to see you here!
\(\displaystyle \L\ \int e^x(1 + x^3)^{\frac{5}{3}} \ dx\) = \(\displaystyle \L\ e^x(1 + x^3)^{\frac{5}{3}} - \frac{5}{6}\ e^x(1 + x^3)^2 + \frac{5}{6}\ e^x + \frac{5}{6}\ \int e^x x^3\ dx\)
\(\displaystyle \L\ \int e^x x^3\ dx = e^x(x^3 - 3x^2 + 6x - 6)\)
\(\displaystyle \L\ \int e^x x^3\ dx = e^x(x^3 - 3x^2 + 6x - 6)\)