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Question Number 169050 by Mastermind last updated on 23/Apr/22

Solve the ODE y′ + xy = x^2 , with y(0)=2 Mastermind

$${Solve}\:{the}\:{ODE} \\ $$$${y}'\:+\:{xy}\:=\:{x}^{\mathrm{2}} ,\:{with}\:{y}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by Mathspace last updated on 23/Apr/22

h→y^′ =−xy ⇒(y^′ /y)=−x ⇒ ln∣y∣=−(x^2 /2)+λ ⇒y=k e^(−(x^2 /2)) (mvc)→y^′ =k^′ e^(−(x^2 /2)) −kx e^(−(x^2 /2)) (e)⇒k^(′ ) e^(−(x^2 /2)) −kx e^(−(x^2 /2)) +xk e^(−(x^2 /2)) =x^2 ⇒k^′ =x^2 e^(x^2 /2) ⇒k=∫_0 ^x t^2 e^(t^2 /2) dt +λ y(x)=(λ+∫_0 ^x t^2 e^(t^2 /2) dt)e^(−(x^2 /2)) y(0)=2 ⇒λ=2 ⇒ y(x)=2e^(−(x^2 /2)) +e^(−(x^2 /2)) ∫_0 ^x t^2 e^(t^2 /2) dt

$${h}\rightarrow{y}^{'} =−{xy}\:\Rightarrow\frac{{y}^{'} }{{y}}=−{x}\:\Rightarrow \\ $$$${ln}\mid{y}\mid=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\lambda\:\Rightarrow{y}={k}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\left({mvc}\right)\rightarrow{y}^{'} ={k}^{'} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} −{kx}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\left({e}\right)\Rightarrow{k}^{'\:} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} −{kx}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{xk}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow{k}^{'} ={x}^{\mathrm{2}} \:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\Rightarrow{k}=\int_{\mathrm{0}} ^{{x}} \:{t}^{\mathrm{2}} \:{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \:{dt}\:+\lambda \\ $$$${y}\left({x}\right)=\left(\lambda+\int_{\mathrm{0}} ^{{x}} \:{t}^{\mathrm{2}} \:{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \:{dt}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{2}\:\Rightarrow\lambda=\mathrm{2}\:\Rightarrow \\ $$$${y}\left({x}\right)=\mathrm{2}{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:+{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\int_{\mathrm{0}} ^{{x}} \:{t}^{\mathrm{2}} \:{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \:{dt} \\ $$

Commented by Mastermind last updated on 24/Apr/22

I did not understand this sir, could you solve it in another way

$${I}\:{did}\:{not}\:{understand}\:{this}\:{sir}, \\ $$$${could}\:{you}\:{solve}\:{it}\:{in}\:{another}\:{way} \\ $$$$ \\ $$

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