What is a solution to the differential equation dydx=2y+x2x?

This is a linear differential equation. Its solution can be composed of the homogeneous solution (yh(x)) plus a particular solution (yp(x)) soy(x)=yh(x)+yp(x)The homogeneous solution obeysdyhdx−2xyh(x)=0. Here, clearly the solution isyh(x)=Cx2By grouping variables:dyhyh=2dxx→ln⁡yh=C1ln⁡x2→yh=Cx2The determination of yp(x) must obeydypdx−2xyp(x)=xWe will suppose that yp(x)=c(x)yh(x). Substituting we obtainyh(x)dc(x)dx+c(x)(dyh(x)dx−2yh(x)x)=x oryh(x)dc(x)dx=x sodc(x)dx=xyh(x)=xx2=1x soc(x)=logex and finallyy(x)=Cx2+loge(x)x2=(C+loge(x))x2

Alternative Approachy′=2y+x2x=2yx+xy′−2xy=xThis is exact if we times it by a factor of 1x21x2y′−2x3y=1xie(1x2y)′=1x1x2y=ln⁡x+Cy=x2(ln⁡x+C)Integrating FactorWe can extract the "integrating factor" η(x) for the DE in form y′−2xy=x as:η=exp⁡(∫−2xdx)=e−2ln⁡x=1x2

This is a linear differential equation. Its solution can be composed of the homogeneous solution (yh(x)) plus a particular solution (yp(x)) soy(x)=yh(x)+yp(x)The homogeneous solution obeysdyhdx−2xyh(x)=0. Here, clearly the solution isyh(x)=Cx2By grouping variables:dyhyh=2dxx→ln⁡yh=C1ln⁡x2→yh=Cx2The determination of yp(x) must obeydypdx−2xyp(x)=xWe will suppose that yp(x)=c(x)yh(x). Substituting we obtainyh(x)dc(x)dx+c(x)(dyh(x)dx−2yh(x)x)=x oryh(x)dc(x)dx=x sodc(x)dx=xyh(x)=xx2=1x soc(x)=logex and finallyy(x)=Cx2+loge(x)x2=(C+loge(x))x2

Alternative Approachy′=2y+x2x=2yx+xy′−2xy=xThis is exact if we times it by a factor of 1x21x2y′−2x3y=1xie(1x2y)′=1x1x2y=ln⁡x+Cy=x2(ln⁡x+C)Integrating FactorWe can extract the "integrating factor" η(x) for the DE in form y′−2xy=x as:η=exp⁡(∫−2xdx)=e−2ln⁡x=1x2

Answered: "What is a solution to the differential equation..."

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Willow Cooper

Answered question

2022-04-19

What is a solution to the differential equation

\frac{d y}{d x} = \frac{2 y + x^{2}}{x}

Answer & Explanation

veselrompoikm

Beginner2022-04-20Added 12 answers

This is a linear differential equation. Its solution can be composed of the homogeneous solution

(y_{h} (x))

plus a particular solution

(y_{p} (x))

y (x) = y_{h} (x) + y_{p} (x)

The homogeneous solution obeys

\frac{{d y}_{h}}{d x} - \frac{2}{x} y_{h} (x) = 0

. Here, clearly the solution is

y_{h} (x) = C x^{2}

By grouping variables:

\frac{{d y}_{h}}{y_{h}} = 2 \frac{d x}{x} \to {\ln y}_{h} = C_{1} {\ln x}^{2} \to y_{h} = C x^{2}

The determination of

y_{p} (x)

must obey

\frac{{d y}_{p}}{d x} - \frac{2}{x} y_{p} (x) = x

We will suppose that

y_{p} (x) = c (x) y_{h} (x)

. Substituting we obtain

y_{h} (x) \frac{d c (x)}{d x} + c (x) (\frac{{d y}_{h} (x)}{d x} - \frac{2 y_{h} (x)}{x}) = x

y_{h} (x) \frac{d c (x)}{d x} = x

\frac{d c (x)}{d x} = \frac{x}{y_{h} (x)} = \frac{x}{x^{2}} = \frac{1}{x}

c (x) = \log_{e} x

and finally

y (x) = C x^{2} + \log_{e} (x) x^{2} = (C + \log_{e} (x)) x^{2}

abangan85s0

Beginner2022-04-21Added 16 answers

Alternative Approach

y^{'} = \frac{2 y + x^{2}}{x} = 2 \frac{y}{x} + x

y^{'} - \frac{2}{x} y = x

This is exact if we times it by a factor of

\frac{1}{x^{2}}

\frac{1}{x^{2}} y^{'} - \frac{2}{x^{3}} y = \frac{1}{x}

{(\frac{1}{x^{2}} y)}^{'} = \frac{1}{x}

\frac{1}{x^{2}} y = \ln x + C

y = x^{2} (\ln x + C)

Integrating Factor
We can extract the "integrating factor"

η (x)

for the DE in form

y^{'} - \frac{2}{x} y = x

as:

η = \exp (\int - \frac{2}{x} d x)

= e^{- 2 \ln x}

= \frac{1}{x^{2}}

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