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If for x\geqslant 0,y=y(x) is the solution of the differential equation, (x+1)dy=\left ( \left ( x+1 \right )^{2}+y-3 \right )dx,y(2)=0, then y\left ( 3 \right ) is equal to ____________.

 

Option: 1

3


Option: 2

2


Option: 3

-3


Option: 4

-2


Answers (1)

best_answer

 

 

Linear Differential Equation -

The linear differential equations are those in which the variable and its derivative occur only in the first degree.

An equation of the form

\frac{dy}{dx}+P(x)\cdot y=Q(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots\text{(i)}

Where P(x) and Q(x) are functions of x only or constant is called a linear equation of the first order.

To solve the differential equation (i)
\\\text{multiply both sides of Eq (i) by }\int e^{P(x)\;dx}\text{, we get}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}(\mathrm{x}) \mathrm{y}\right)=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}} \cdot \mathrm{Q}(\mathrm{x})\\\\\mathrm{i.e.\;\;\;\;\;\;\;\;e^{\int P(x) d x} \cdot \frac{d y}{d x}+y \frac{d}{d x}\left(e^{\int P(x) d x}\right)=Q e^{\int P(x) d x}}\\\\\mathrm{or\;\;\;\;\;\;\;\;\;\frac{d}{d x}\left(y e^{\int P(x) d x}\right)=e^{\int P(x) d x} \cdot Q(x)}\\\\\text{Integrating both sides, we get}\\\\\mathrm{or\;\;\;\;\;\;\;\;\;\int{d}\left(y e^{\int P(x) d x}\right)=\int\left (e^{\int P(x) d x} \cdot Q(x) \right )dx}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\mathrm{y} \mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}}=\int \mathrm{Q}(\mathrm{x}) \mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}} \mathrm{d} \mathrm{x}+\mathrm{C}}

Which is the required solution of the given differential equation.

 

\\\mathrm{The\;term\;\int e^{P(x)\;dx}\;\;which \;convert\;the\;left\;hand\;expression\;of\;the\;equation}\\\text{into a perfect differential is called an Integrating factor (IF).}\\\\ {\text { Thus, we remember the solution of the above equation as }} \\ {\qquad y(\mathrm{IF})=\int Q(\mathrm{IF}) d x+C}

NOTE : 

Sometimes a given differential equation becomes linear if we take x as the dependent variable and y as the independent variable, i.e. it can be written in the form 

\\\mathrm{The\;term\;\int e^{P(x)\;dx}\;\;which \;convert\;the\;left\;hand\;expression\;of\;the\;equation}\\\text{into a perfect differential is called an Integrating factor (IF).}\\\\ {\text { Thus, we remember the solution of the above equation as }} \\ {\qquad y(\mathrm{IF})=\int Q(\mathrm{IF}) d x+C}

-

(1+x) \frac{d y}{d x}=\left((1+x)^{2}+(y-3)\right)

(1+x) \frac{d y}{d x}-y=(1+x)^{2}-3

\frac{d y}{d x}-\frac{1}{(1+x)} y=(1+x)-\frac{3}{(1+x)}

I F=e^{-\int \frac{1}{1+x} d x}=\frac{1}{1+x}

\frac{d}{d x}\left(\frac{y}{1+x}\right)=1-\frac{3}{(1+x)^{2}}

x=2, y=0, c=-3

So at x=3 y=3

 

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Gunjita

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