InterviewSolution
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InterviewSolution
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Solve `(x^2+1)dy/dx+2xy=sqrt(x^2+4)` |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (( 2x)/( x ^(2) + 1 )) y = ( sqrt ( x ^(2) + 4))/( (x ^(2) + 1)) " " ` ... (i) This is of the form ` (dy)/(dx) + P y = Q,` `" " ` where ` P= (2x)/(( x ^(2) + 1 )) and Q = (sqrt (x ^(2) + 4))/( ( x ^(2) + 1 ))` Thus, the given differential equation is linear. `IF = e ^( int Pdx)= e ^( int (2x)/((x ^(2) + 1 )) dx) = e ^( log (x ^(2) + 1 )) = (x ^(2)+ 1)` So, the required solution is given by `y xx IF = int { Q xx IF} dx + C`, i.e., `y (x^(2) + 1 ) = int(sqrt (x^(2) + 4))/( (x^(2) + 1 )) xx (x ^(2) + 1) dx` `rArr y (x ^(2) + 1 ) = int sqrt ( x^(2) + 4) dx ` ` " "= (1)/(2) x sqrt (x^(2) + 4) + (1)/(2) xx 2^(2) xx log |x + sqrt (x ^(2) + 4)| + C` `" " = (1)/(2) x sqrt (x^(2) + 4) + 2log |x + sqrt (x ^(2) + 4)| + C`. Hence, `y (x ^(2) + 1 ) = (1)/(2) x sqrt ( x ^(2) + 4) + 2log | x + sqrt (x ^(2) + 4) | + C ` is the required solution. |
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